{"id":37822,"date":"2022-12-03T12:21:14","date_gmt":"2022-12-03T06:51:14","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=37822"},"modified":"2022-12-05T11:22:38","modified_gmt":"2022-12-05T05:52:38","slug":"maths-2a-binomial-theorem-important-questions-long-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/maths-2a-binomial-theorem-important-questions-long-answer-type\/","title":{"rendered":"TS Inter Second Year Maths 2A Binomial Theorem Important Questions Long Answer Type"},"content":{"rendered":"

Students must practice these Maths 2A Important Questions<\/a> TS Inter Second Year Maths 2A Binomial Theorem Important Questions Long Answer Type to help strengthen their preparations for exams.<\/p>\n

TS Inter Second Year Maths 2A Binomial Theorem Important Questions Long Answer Type<\/h2>\n

Question 1.
\nState and prove Binomial Theorem. [March ’09]
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the numerically greatest term in the expansion of (4 + 3x)15<\/sup> where x = \\(\\frac{7}{2}\\).
\nSolution:
\nGiven
\n(4 + 3x)15<\/sup>
\n\u21d2 415<\/sup> (1 + \\(\\frac{3 x}{4}\\))15<\/sup> = 415<\/sup> (1 + x)n<\/sup>
\nwhere n = 15; x = \\(\\frac{3 x}{4}\\)
\n\\(|x|=\\left|\\frac{3 x}{4}\\right|=\\left|\\frac{3}{4} \\cdot \\frac{7}{2}\\right|=\\frac{21}{8}\\)
\nNow, \\(\\frac{(n+1)|x|}{|x|+1}=\\frac{(15+1)\\left|\\frac{3 x}{4}\\right|}{\\left|\\frac{3 x}{4}\\right|+1}\\)
\n= \\(\\frac{16 \\cdot \\frac{21}{8}}{\\frac{21}{8}+1}=\\frac{336}{29}\\) = 11.5
\n\u2234 T12<\/sub> is numerically greatest.
\nr + 1 = 12
\n\u21d2 r = 11
\nThe general term in the expansion of (x + a)n<\/sup> is
\nTr + 1<\/sub> = \\({ }^n C_r\\) xn – r<\/sup> ar<\/sup>
\nT11 + 1<\/sub> = \\({ }^{15} \\mathrm{C}_{11}\\) (4)15 – 11<\/sup> (3x)11<\/sup>
\n12th term in this expansion is<\/p>\n

\"TS<\/p>\n

Question 3.
\nFind the numerically reatest tenu in the expansion of (4a – 6b)13<\/sup> when a = 3, b = 5.
\nSolution:
\nGiven (4a – 6b)13<\/sup> = (4a)13<\/sup> (1 – \\(\\frac{6 b}{4 a}\\))13<\/sup>
\n= (4a)13<\/sup> (1 + x)n<\/sup>
\nwhere, x = \\(-\\frac{6 b}{4 a}\\), n = 13
\n|x| = \\(\\left|\\frac{-6 b}{4 a}\\right|=\\left|\\frac{-6 \\cdot 5}{4 \\cdot 3}\\right|=\\frac{5}{2}\\)
\nNow, \\(\\frac{(n+1)|x|}{|x|+1}=\\frac{(13+1) \\frac{5}{2}}{\\frac{5}{2}+1}\\)
\n= \\(\\frac{14 \\cdot 5}{5+2}=\\frac{70}{7}\\) = 10
\n\u2234 T10<\/sub> and T11<\/sub> are numerically greatest.
\nT10<\/sub>:
\nr + 1 = 10
\n\u21d2 r = 9
\nThe general term in the expansion of (1 + x)n<\/sup> is
\nTr + 1<\/sub> = \\({ }^{\\mathrm{n}} \\mathrm{C}_{\\mathrm{r}}\\) xn – r<\/sup> ar<\/sup>
\nT10<\/sub> in this expansion is
\nT9+1<\/sub> = \\({ }^{13} \\mathrm{C}_9\\) (4a)13-9<\/sup> (- 6b)9<\/sup>
\nT10 = \\({ }^{13} \\mathrm{C}_9\\) (4a)4<\/sup> (- 6b)9<\/sup>
\n= \\({ }^{13} \\mathrm{C}_9\\) (44<\/sup> . 34<\/sup>) ((- 6)9<\/sup> (5)9<\/sup>)
\n= \\({ }^{13} \\mathrm{C}_9\\) 124<\/sup> . 309<\/sup>
\n|T10<\/sub>| = \\({ }^{13} \\mathrm{C}_9\\) 124<\/sup> . 309<\/sup><\/p>\n

T11<\/sub>:
\nr + 1 = 11
\n\u21d2 r = 10.
\nThe general term in the expansion of (1 – x)n<\/sup> is
\nTr + 1<\/sub> = \\({ }^{\\mathrm{n}} \\mathrm{C}_{\\mathrm{r}}\\) xn – r<\/sup> ar<\/sup>
\nT11<\/sub> in this expansion is
\nT10+1<\/sub> = \\({ }^{13} \\mathrm{C}_{10}\\) (4a)13-10<\/sup> . (- 6b)10<\/sup>
\n= \\({ }^{13} \\mathrm{C}_{10}\\) (4a)3<\/sup> (6b)10<\/sup>
\nT11<\/sub> = \\({ }^{13} \\mathrm{C}_{10}\\) (4 . 3)3<\/sup> (6 . 5)10<\/sup>
\n= \\({ }^{13} \\mathrm{C}_{10}\\) 123<\/sup> . 3010<\/sup>
\nT11<\/sub> = \\({ }^{13} \\mathrm{C}_{10}\\) 123<\/sup> . 3010<\/sup>
\n\u2234 |T10<\/sub>| = |T11<\/sub>|.<\/p>\n

\"TS<\/p>\n

Question 4.
\nIf the coefficient of x10<\/sup> in the expansion of \\(\\left(a x^2+\\frac{1}{b x}\\right)^{11}\\) is equal to the coefficient of x-10<\/sup> in the expansion of \\(\\left(a x-\\frac{1}{b x^2}\\right)^{11}\\) find the relation between a and b where a and b are real numbers. [AP – May 2015, AP – Mar. 2019]
\nSolution:
\nCase – I :
\nGiven \\(\\left(a x^2+\\frac{1}{b x}\\right)^{11}\\)
\nHere x = ax2<\/sup>, a = \\(\\frac{1}{\\mathrm{bx}}\\); n = 11
\nNow, the general term in the expansion is
\nTr+1<\/sub> = \\({ }^n C_r\\) xn-r<\/sup> ar<\/sup>
\n= \\({ }^{11} \\mathrm{C}_{\\mathrm{r}}\\) a11-r<\/sup> x22-2r<\/sup> 1r<\/sup> b-r<\/sup> x-r<\/sup>
\n= \\({ }^{11} \\mathrm{C}_{\\mathrm{r}}\\) a11-r<\/sup> b-r<\/sup> x22-3r<\/sup> …………….(1)
\nTo find the coefficient of x10<\/sup>,
\nPut 22 – 3r = 10
\n3r = 12
\nr = 4
\nSubstituting r = 4 in equation (1) we get
\nT4+1<\/sub> = \\({ }^{11} C_4\\) a11-4<\/sup> b-4<\/sup> x22-12<\/sup>
\nT5<\/sub> = \\({ }^{11} C_4\\) a7<\/sup> b-4<\/sup> x10<\/sup>
\nT5<\/sub> = \\({ }^{11} C_4\\) a7<\/sup> b-4<\/sup> x10<\/sup>
\n\u2234 The coeff. of x10<\/sup> \u00a1n the expansion of \\(\\left(a x^2+\\frac{1}{b x}\\right)^{11}\\) is \\({ }^{11} C_4\\) a7<\/sup> b-4<\/sup><\/p>\n

Case II:
\nGiven \\(\\left(a x-\\frac{1}{b x^2}\\right)^{11}\\)
\nHere x = ax, a = \\(\\left(\\frac{-1}{b x^2}\\right)\\), n = 11
\nNow, the general term in the expansion is
\nTr + 1<\/sub> = \\({ }^n C_r\\) xn-r<\/sup> ar<\/sup>
\n= \\({ }^{11} C_r(a x)^{11-r}\\left(\\frac{-1}{b x^2}\\right)^r\\)
\n= \\({ }^{11} \\mathrm{C}_{\\mathrm{r}}\\) a11-r<\/sup> x11-r<\/sup> (- 1)r<\/sup> b-r<\/sup> x-2r<\/sup>
\n= \\({ }^{11} \\mathrm{C}_{\\mathrm{r}}\\) (- 1)r<\/sup> b-r<\/sup> x11-3r<\/sup> ……………(2)
\nTo find the coefficient of x-10<\/sup>
\nput 11 – 3r = – 10
\n\u21d2 3r = 21
\n\u21d2 r = 7
\nSubstitute r = 7 in equation (2) we get
\nT7+1<\/sub> = \\({ }^{11} \\mathrm{C}_7\\) a11-7<\/sup> b7<\/sup> (- 1)7<\/sup> x11-21<\/sup>
\nT8<\/sub> = – \\({ }^{11} \\mathrm{C}_7\\) a4<\/sup> b-7<\/sup> x-10<\/sup>
\n\u2234 The coeff. of x-10<\/sup> in the expansion of \\(\\left(a x-\\frac{1}{b x^2}\\right)^{11}\\) is – \\({ }^{11} \\mathrm{C}_7\\) a4<\/sup> b-7<\/sup>
\nGiven that, these coefficients are equal
\n\\({ }^{11} \\mathrm{C}_7\\) a7<\/sup> b-4<\/sup> = – \\({ }^{11} \\mathrm{C}_7\\) a4<\/sup> b-7<\/sup>
\n\\({ }^{11} C_4 \\frac{a^7}{b^4}=-{ }^{11} C_4 \\frac{a^4}{b^7}\\)
\n\\(\\frac{a^7}{b^4}=\\frac{-a^4}{b^7}\\)
\n\u21d2 a3<\/sup>b3<\/sup> = – 1
\n\u21d2 (ab)3<\/sup> = (- 1)3<\/sup>
\n\u21d2 ab = – 1<\/p>\n

Question 5.
\nIf n is a positive integer, then show that
\ni) C0<\/sub> + C1<\/sub> + C2<\/sub> + ……………… + Cn<\/sub> = 2
\nii) C0<\/sub> + C2<\/sub> + C4<\/sub> + ……………. = C1<\/sub> + C3<\/sub> + C5<\/sub> + ……….. = 2n-1<\/sup> [March ’97, ’90]
\nSolution:
\nWe know that
\n(1 + x)n<\/sup> = C0<\/sub> + C1<\/sub>x + C2<\/sub>x2<\/sup> + ……………… + Cn<\/sub>xn<\/sup> ………….(1)
\ni) Put x = 1 in equation (1)
\n\u21d2 (1 + 1)n<\/sup> = C0<\/sub> + C1<\/sub>(1) + C2<\/sub>(1)2<\/sup> + C3<\/sub>(1)3<\/sup> + C4<\/sub>(1)4<\/sup> + ……………… + Cn<\/sub>(1)n<\/sup>
\n2n<\/sup> = C0<\/sub> + C1<\/sub> + C2<\/sub> + C3<\/sub> + C4<\/sub> + …………. + Cn<\/sub>
\n\u2234 C0<\/sub> + C1<\/sub> + C2<\/sub> + C3<\/sub> + C4<\/sub> + …………. + Cn<\/sub> = 2n<\/sup><\/p>\n

ii) Put x = – 1 in equation (1), we get
\n(1 – 1)n<\/sup> = C0<\/sub> + C1<\/sub> (- 1) + C2<\/sub> (- 1)2<\/sup> + C3<\/sub> (- 1)3<\/sup> + ……………. + Cn<\/sub> (- 1)n<\/sup>
\no = C0<\/sub> – C1<\/sub> + C2<\/sub> – C3<\/sub> + C4<\/sub> – ………………
\nC0<\/sub> + C2<\/sub> + C4<\/sub> + …………………. = C1<\/sub> + C3<\/sub> + C5<\/sub> + ………….
\nSince a = b
\n\u21d2 a = b = \\(\\frac{a+b}{2}\\)
\nC0<\/sub> + C2<\/sub> + C4<\/sub> + …………………. = C1<\/sub> + C3<\/sub> + C5<\/sub> + ………….
\n= \\(\\frac{\\mathrm{C}_0+\\mathrm{C}_2+\\mathrm{C}_4+\\ldots \\ldots+\\mathrm{C}_1+\\mathrm{C}_3+\\mathrm{C}_5+\\ldots \\ldots}{2}\\)
\n= \\(\\frac{\\mathrm{C}_0+\\mathrm{C}_1+\\mathrm{C}_2+\\mathrm{C}_3+\\mathrm{C}_4+\\mathrm{C}_5+\\ldots \\ldots}{2}\\)
\n= \\(\\frac{2^{\\mathrm{n}}}{2}\\)
\n= 2n – 1<\/sup>
\n\u2234 C0<\/sub> + C2<\/sub> + C4<\/sub> + …………………. = C1<\/sub> + C3<\/sub> + C5<\/sub> + …………. = 2n – 1<\/sup>
\nC0<\/sub> + C2<\/sub> + C4<\/sub> + …………………. = 2n – 1<\/sup>
\nC1<\/sub> + C3<\/sub> + C5<\/sub> + …………………… = 2n – 1<\/sup><\/p>\n

\"TS<\/p>\n

Question 6.
\nProve that for any real numbers a, d, a . C0<\/sub> + (a + d) . C1<\/sub> + (a + 2d) . C2<\/sub> + ……………… + (a + nd) . Cn<\/sub> = (2a + nd) 2n – 1<\/sup> [May \u201898]
\nSolution:
\nLet S = a . C0<\/sub> + (a + d) . C1<\/sub> + (a + 2d) . C2<\/sub> + ……………… + (a + nd) . Cn<\/sub> ………………(1)
\nBy writing the terms in R.H.S of (1) in reverse order has done we get
\nS = (a + nd)Cn<\/sub> + (a + (n – 1)d)Cn-1<\/sub> + (a + (n-2)d)Cn-2<\/sub> + …………. + aC0<\/sub>
\nS = (a + nd)C0<\/sub> + (a + (n – 1)d)C1<\/sub> + (a + (2n – 2)d)C2<\/sub> + ………….. + aCn<\/sub> ……………(2)
\nAdding (1) and (2) we get<\/p>\n

\"TS<\/p>\n

2s = (2a + nd) (C0<\/sub> + C1<\/sub> + C2<\/sub> + …………. + Cn<\/sub>)
\n2s = (2a + nd)2n<\/sup>
\n\u21d2 S = (2a + nd)2n-1<\/sup>
\n\u2234 aC0<\/sub> + (a + d)C1<\/sub> + (a + 2d)C2<\/sub> + …………. + (a + nd)Cn<\/sub> = (2a + nd) 22n-1<\/sup>.<\/p>\n

Question 7.
\nFor r = 0, 1, 2, ….., n, prove that C0<\/sub> . Cr<\/sub> + C1<\/sub> . Cr+1<\/sub> + C2<\/sub> . Cr+2<\/sub> + ……………… + Cn-r<\/sub> . Cn<\/sub> = \\({ }^{2 n} C_{(n+r)}\\)
\nand hence deduce that
\ni) C0<\/sub>2<\/sup> + C1<\/sub>2<\/sup> + C2<\/sub>2<\/sup> + ………….. + Cn<\/sub>2<\/sup> = \\({ }^{2 n} C_n\\)
\nii) C0<\/sub> . C1<\/sub> + C1<\/sub> . C2<\/sub> + C2<\/sub> . C3<\/sub> + ………… + Cn-1<\/sub> . Cn<\/sub> = \\({ }^{2 n} C_{n+1}\\)
\n[May \u201897, 95, \u201890, \u201894,\u201993, \u201892; March \u201898, \u201893, AP-Mar. 2018; TS – Mar. 2015]
\nSolution:
\nWe know that
\n(1 + x)n<\/sup> = C0<\/sub> + C1<\/sub> . x + C2<\/sub> . x2<\/sup> + ………………….. + Cr<\/sub> xr<\/sup> + Cn<\/sub> . xn<\/sup> ………………(1)
\n(x + 1)n<\/sup> = C0<\/sub>xn<\/sup> + C1<\/sub>xn-1<\/sup> + C2<\/sub>xn-2<\/sup> + ………………… + Cr<\/sub>xn-r<\/sup> + …………………. + Cn<\/sub> ………………(2)
\nMultiplying (2) and (1), we get
\n(C0<\/sub>xn<\/sup> + C1<\/sub>xn-1<\/sup> + C2<\/sub>xn-2<\/sup> + ………………… + Cr<\/sub>xn-r<\/sup> + …………………. + Cn<\/sub>) (C0<\/sub> + C1<\/sub> . x + C2<\/sub> . x2<\/sup> + ………………….. + Cr<\/sub> xr<\/sup> + Cn<\/sub> . xn<\/sup>)
\n= (x + 1)n<\/sup> (1 + x)n<\/sup> = (1 + x)2n<\/sup>
\nComparing the coefficient of xn+r<\/sup> on bothsides we get,
\nC0<\/sub> . Cr<\/sub> + C1<\/sub> . Cr+1<\/sub> + C2<\/sub> . Cr+2<\/sub> + ……………… + Cn-r<\/sub> . Cn<\/sub> = \\({ }^{2 n} C_{(n+r)}\\) ……………….(3)<\/p>\n

i) On substituting r = 0 in equation (3) we get
\nC0<\/sub> . C0<\/sub> + C1<\/sub> . C1<\/sub> + C2<\/sub> . C2<\/sub> + ……………….. + Cn<\/sub> . Cn<\/sub> = \\({ }^{2 n} C_n\\)
\nC0<\/sub>2<\/sup> + C1<\/sub>2<\/sup> + C2<\/sub>2<\/sup> + ………….. + Cn<\/sub>2<\/sup> = \\({ }^{2 n} C_n\\)<\/p>\n

ii) On substituting, r = 1 in equation (3) we get
\nC0<\/sub> . C1<\/sub> + C1<\/sub> . C2<\/sub> + C2<\/sub> . C3<\/sub> + ………… + Cn-1<\/sub> . Cn<\/sub> = \\({ }^{2 n} C_{n+1}\\)<\/p>\n

Question 8.
\nIf n is a positive integer and x is any non-zero real number, then prove that
\nC0<\/sub> + C1<\/sub> . \\(\\frac{x}{2}\\) + C2<\/sub> . \\(\\frac{x^2}{3}\\) + C3<\/sub> . \\(\\frac{x^3}{4}\\) + …………….. + Cn<\/sub> . \\(\\frac{\\mathbf{x}^n}{n+1}\\) = \\(\\frac{(1+x)^{n+1}-1}{(n+1) x}\\). [May ’14, ’13, ’08, ’05, Mar. ’14, ’02, Board Paper, AP – May 2016]
\nSolution:
\nLet S = C0<\/sub> + C1<\/sub> . \\(\\frac{x}{2}\\) + C2<\/sub> . \\(\\frac{x^2}{3}\\) + C3<\/sub> . \\(\\frac{x^3}{4}\\) + …………….. + Cn<\/sub> . \\(\\frac{\\mathbf{x}^n}{n+1}\\) = \\(\\frac{(1+x)^{n+1}-1}{(n+1) x}\\)<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 9.
\nIf n is a positive integer, then prove that
\nC0<\/sub> + \\(\\frac{C_1}{2}+\\frac{C_2}{3}+\\ldots .+\\frac{C_n}{n+1}=\\frac{2^{n+1}-1}{n+1}\\) [TS – Mar. 2017]
\nSolution:
\nLet S = C0<\/sub> + \\(\\frac{\\mathrm{C}_1}{2}+\\frac{\\mathrm{C}_2}{3}+\\ldots . .+\\frac{\\mathrm{C}_{\\mathrm{n}}}{\\mathrm{n}+1}\\)
\n= \\({ }^{\\mathrm{n}} \\mathrm{C}_0+\\frac{1}{2} \\cdot{ }^{\\mathrm{n}} \\mathrm{C}_1+\\frac{1}{3} \\cdot{ }^{\\mathrm{n}} \\mathrm{C}_2+\\frac{1}{\\mathrm{n}+1} \\cdot{ }^{\\mathrm{n}} \\mathrm{C}_{\\mathrm{n}}\\)
\nNow multiplying on both sides with (n + 1), we get<\/p>\n

\"TS<\/p>\n

Question 10.
\nProve that \\(\\frac{C_1}{2}+\\frac{C_3}{4}+\\frac{C_5}{6}+\\frac{C_7}{8}+\\cdots \\cdots=\\frac{2^n-1}{n+1}\\) [May 08].
\nSolution:<\/p>\n

\"TS<\/p>\n

Question 11.
\nIf (1 + x + x2<\/sup>)n<\/sup> = a0<\/sub> + a1<\/sub>x + a2<\/sub>x2<\/sup> + ……………… + a2n<\/sub>x2n<\/sup>, then prove that
\ni) a0<\/sub> + a1<\/sub> + a2<\/sub> + …………….. + a2n<\/sub> = 3n<\/sup>
\nii) a0<\/sub> + a2<\/sub> + a4<\/sub> + ……………. + a2n<\/sub> = \\(\\frac{3^n+1}{2}\\)
\niii) a1<\/sub> + a3<\/sub> + a5<\/sub> + ……………… + a2n<\/sub> = \\(\\frac{3^n-1}{2}\\)
\niv) a0<\/sub> + a3<\/sub> + a6<\/sub> + a9<\/sub> + ………………. = 3n-1<\/sup> [TS May 2016; Board Paper]
\nSolution:
\n(1 + x + x2<\/sup>)n<\/sup> = a0<\/sub> + a1<\/sub>x + a2<\/sub>x2<\/sup> + ……………… + a2n<\/sub>x2n<\/sup>
\nPut x = 1 in equation (1) we get,
\n(1 + 1 + 12<\/sup>)n<\/sup> = a0<\/sub> + a1<\/sub> (1) + a2<\/sub> (1)2<\/sup> + ……………… + a2n<\/sub>. (1)2n<\/sup>
\na0<\/sub> + a1<\/sub> + a2<\/sub> + …………… + a2n<\/sub> = 3n<\/sup> ……………(2)
\nPut x = – 1 in equation (1) we get,
\n(1 – 1 + 12<\/sup>)n<\/sup> = a0<\/sub> + a1<\/sub> (- 1) + a2<\/sub> (- )2<\/sup> + a3<\/sub>. (- 1)4<\/sup> ……………… + a2n<\/sub>. (- 1)2n<\/sup>
\na0<\/sub> – a1<\/sub> + a2<\/sub> – a3<\/sub> + a4<\/sub> + …………… + a2n<\/sub> = 1 ……………(3)
\ni) From (2),
\na0<\/sub> + a1<\/sub> + a2<\/sub> + a3<\/sub> + …………… + a2n<\/sub> = 3n<\/sup><\/p>\n

ii) Now, adding (2) and (3)
\na0<\/sub> + a1<\/sub> + a2<\/sub> + a3<\/sub> + …………… + a2n<\/sub> = 3n<\/sup><\/p>\n

\"TS<\/p>\n

iii) Now, subtracting (2) and (3)<\/p>\n

\"TS<\/p>\n

iv) Put, x = 1 in equation (1) we get
\n(1 + 1 + 12<\/sup>)n<\/sup> = a0<\/sub> + a1<\/sub> (1) + a2<\/sub>(1)2<\/sup> + a3<\/sub> (1)3<\/sup> + a4<\/sub> (1)4<\/sup> + a5<\/sub> (1)5<\/sup> + a6<\/sub> (1)6<\/sup> + ……………… + a2n<\/sub> (1)2n<\/sup>
\n= a0<\/sub> + a1<\/sub> + a2<\/sub> + a3<\/sub> + a4<\/sub> + a5<\/sub> + a6<\/sub> + …………….. + a2n<\/sub> = 3n<\/sup> ………………(4)<\/p>\n

Put x = \u03c9 in equation (1), we get
\n(1 + \u03c9 + \u03c92<\/sup>)n<\/sup> = a0<\/sub> + a1<\/sub> (\u03c9) + a2<\/sub> (\u03c9)2<\/sup> + a3<\/sub> (\u03c9)3<\/sup> + a4<\/sub> (\u03c9)4<\/sup> + a5<\/sub> (\u03c9)5<\/sup>+ a6<\/sub> (\u03c9)6<\/sup> + ……………. + a2n<\/sub> (\u03c9)2n<\/sup>
\n= a0<\/sub> + a1<\/sub> . (\u03c9) + a2<\/sub> (\u03c9)2<\/sup> + a3<\/sub> (\u03c9)3<\/sup> + a4<\/sub> (\u03c9)4<\/sup> + a5<\/sub> (\u03c9)5<\/sup> + a6<\/sub> (\u03c9)6<\/sup> + ……………. + a2n<\/sub> (\u03c9)2n<\/sup> = 0 …………….(5)
\nput x = in equation (1) we get,
\n(1 + \u03c92<\/sup> + \u03c94<\/sup>)n<\/sup> = a0<\/sub> + a1<\/sub> . (\u03c9)2<\/sup> + a2<\/sub> (\u03c9)4<\/sup> + a3<\/sub> (\u03c9)6<\/sup> + a4<\/sub> (\u03c9)8<\/sup> + a5<\/sub> (\u03c9)10<\/sup> + a6<\/sub> (\u03c9)12<\/sup> + ……………. + a2n<\/sub> (\u03c92<\/sup>)2n<\/sup>
\n= a0<\/sub> + a1<\/sub> . \u03c92<\/sup> + a2<\/sub> \u03c9 + a3<\/sub> + a4<\/sub> \u03c92<\/sup> + a5<\/sub> \u03c9 + a6<\/sub> + ……………. + a2n<\/sub> (\u03c92<\/sup>)2n<\/sup> = 0
\nNow, adding (4), (5) and (6)<\/p>\n

\"TS<\/p>\n

\u21d2 3a0<\/sub> + 0 + 0 + 3a3<\/sub> + 0 + 0 + 3a6<\/sub> + ………….. = 3n<\/sup>
\n\u21d2 3(a0<\/sub> + a3<\/sub> + a6<\/sub> + ………………..) = 3n<\/sup>
\n\u21d2 a0<\/sub> + a3<\/sub> + a6<\/sub> + ……………….. = 3n-1.<\/sup><\/p>\n

\"TS<\/p>\n

Question 12.
\nIf (1 + x + x2<\/sup> + x3<\/sup>)7<\/sup> = b0<\/sub> + b1<\/sub>x + b2<\/sub>x2<\/sup> + ………………. + b21<\/sub>x21<\/sup>, then find the value of
\ni) b0<\/sub> + b2<\/sub> + b4<\/sub> + …………. + b20<\/sub>
\nii) b1<\/sub> + b3<\/sub> + b5<\/sub> + …………….. + b21<\/sub>
\nSolution:
\nGiven
\n(1 + x + x2<\/sup> + x3<\/sup>)7<\/sup> = b0<\/sub> + b1<\/sub>x + b2<\/sub>x2<\/sup> + ………………. + b21<\/sub>x21<\/sup> …………………….(1)
\nSubstituting x = 1 in (1),
\nwe get b0<\/sub> + b1<\/sub> + b2<\/sub> + …………. + b21<\/sub> = 47 ………………(2)
\nSubstituting x = – 1 in (1),
\nwe get b0<\/sub> – b1<\/sub> + b2<\/sub> + ……………… – b21<\/sub> = 0 ……………..(3)
\ni) (2) + (3)
\n\u21d2 2b0<\/sub> + 2b2<\/sub> + 2b4<\/sub> + ………………….. + 2b90<\/sub> = 47<\/sup>
\n\u21d2 b0<\/sub> + b2<\/sub> + b4<\/sub> + …………….. + b20<\/sub> = 213<\/sup><\/p>\n

ii) (2) – (3)
\n\u21d2 2b1<\/sub> + 2b3<\/sub> + 2b5<\/sub> + ……………. + 2b21<\/sub> = 47<\/sup>
\n\u21d2 b1<\/sub> + b3<\/sub> + b5<\/sub> + ……………. + b21<\/sub> = 213<\/sup><\/p>\n

Question 13.
\nthe coefficients of x9<\/sup>, x10<\/sup>, x11<\/sup> in the expansion of (1 + x)n<\/sup> are in A.P., then prove that n2<\/sup> – 41n + 398 = 0.
\nSolution:
\nCoefficient of r in the expansion of (1 + x)n<\/sup> is \\({ }^n C_r\\).
\nGiven coefficients of x9<\/sup>, x10<\/sup>, x11<\/sup> in the expansion of (1 + x)n<\/sup> are \\({ }^n C_9,{ }^n C_{10},{ }^n C_{11}\\) in AP., then 2\\(\\left({ }^n C_{10}\\right)={ }^n C_9+{ }^n C_{11}\\)
\n\u21d2 \\(2 \\frac{n !}{(n-10) ! 10 !}=\\frac{n !}{(n-9) ! 9 !}+\\frac{n !}{(n-11) !+11 !}\\)
\n\u21d2 \\(\\frac{2}{10(n-10)}=\\frac{1}{(n-9)(n-10)}+\\frac{1}{11 \\times 10}\\)
\n\u21d2 \\(\\frac{2}{(n-10) 10}=\\frac{110+(n-9)(n-10)}{110(n-9)(n-10)}\\)
\n\u21d2 22 (n – 9) = 110 + n2<\/sup> – 19n + 90
\n\u21d2 n2<\/sup> – 41n + 398 = 0.<\/p>\n

Question 14.
\n1f 36, 84, 126 are three successive binomial coefficlen\u00fc* in the expansion of (1 + x)n<\/sup>, then find n. [May ’09, ’06] [TS – Mar. 2019]
\nSolution:
\nWe know that
\n(1 + x)n<\/sup> = C0<\/sub> + C1<\/sub>x + C2<\/sub>x2<\/sup> + ………………. Cn<\/sub>xn<\/sup>.
\nLet the three successive binomial coefficients in the expansion of (1 + x)n<\/sup> are \\({ }^{\\mathrm{n}} \\mathrm{C}_{\\mathrm{r}-1},{ }^{\\mathrm{n}} \\mathrm{C}_{\\mathrm{r}},{ }^{\\mathrm{n}} \\mathrm{C}_{\\mathrm{r}+1}\\)
\nGiven that,
\n\\({ }^{\\mathrm{n}} \\mathrm{C}_{\\mathrm{r}-1}\\) = 36 ………………(1)
\n\\({ }^{\\mathrm{n}} \\mathrm{C}_{\\mathrm{r}}\\) = 84 ………………(2)
\n\\({ }^{\\mathrm{n}} \\mathrm{C}_{\\mathrm{r}+1}\\) = 126 ……………..(3)<\/p>\n

\"TS<\/p>\n

\u21d2 2n – 2r = 3r + 3
\n\u21d2 5r = 2n – 3 from (4)
\n\u21d2 5 \\(\\left(\\frac{3 n+3}{10}\\right)\\) = 2n – 3
\n\u21d2 3n + 3 = 4n – 6
\n\u21d2 n = 9.<\/p>\n

\"TS<\/p>\n

Question 15.
\nIf the 2nd, 3rd and 4th terms in the expansion of (a + x)n<\/sup> are respectively 240, 720, 1080, find a, x, n. [TS – Mar. 2016; Ma \u201809, ’06]
\nSolution:
\nThe general term in the expansion of (x + a)n<\/sup> is
\nTr+1<\/sub> = xn-r<\/sup> . ar<\/sup>
\nIn the expansion, (a + x)n<\/sup>
\nT2<\/sub> = T1+1<\/sub>
\n= \\({ }^n C_1\\) (a)n-1<\/sup> x1<\/sup> = 240 …………..(1)
\nT3<\/sub> = T2+1<\/sub>
\n= \\({ }^n C_2\\) (a)n-2<\/sup> x2<\/sup> = 720 …………..(2)
\nT4<\/sub> = T3+1<\/sub>
\n= \\({ }^n C_3\\) (a)n-3<\/sup> x3<\/sup> = 1080 …………..(3)<\/p>\n

\"TS<\/p>\n

\u21d2 2(n – 2) x = 9a ………………(5)
\n\\(\\frac{(5)}{(4)}=\\frac{2(n-2) x}{(n-1) x}=\\frac{9 a}{6 a}\\)
\n\u21d2 4(n – 2) = 3(n – 1)
\n\u21d2 4n – 8 = 3n – 3
\n\u21d2 n = 5
\nSubstitute n = 5 in equation (4)
\n\u21d2 (5 – 1)x = 6a
\n\u21d2 4x = 6a
\n\u21d2 2x = 3a
\n\u21d2 x = \\(\\frac{3 a}{2}\\)
\nSubstitute n, x in equation (1).
\n\\(5 C_1(a)^{5-1}\\left(\\frac{3 a}{2}\\right)^1\\) = 240
\n\u21d2 5 . a4<\/sup> . \\(\\frac{3}{2}\\) . a = 240
\n\u21d2 a5<\/sup> = 32
\n\u21d2 a = 2.
\nx = \\(\\frac{3 a}{2}\\)
\nx = 3 . \\(\\frac{2}{2}\\)
\n\u21d2 x = 3
\n\u2234 a = 2, x = 3, n = 5.<\/p>\n

Question 16.
\nIf the coefficients of rth, (r + 1)th and (r + 2)nd tenus in the expansion of (1 + x)n<\/sup> are in A.P then show that n2<\/sup> – (4r + 1) n + 4r2<\/sup> – 2 = 0.
\nSolution:
\nIn the expansion (1 + x)n<\/sup>
\nThe coeff. of Tr<\/sub> = \\({ }^n \\mathrm{C}_{\\mathrm{r}-1}\\);
\nThe coeff. of Tr+1<\/sub> = \\({ }^n C_r\\)
\nThe coe\u00cdf. of Tr+2<\/sub> = \\({ }^n \\mathrm{C}_{\\mathrm{r}+1}\\)
\nGiven that
\n\\({ }^n \\mathrm{C}_{\\mathrm{r}-1}\\), \\({ }^n C_r\\), \\({ }^n \\mathrm{C}_{\\mathrm{r}+1}\\) are in AP.
\nIf a, b, c are in A.P.
\nThen 2b = a + c<\/p>\n

\"TS<\/p>\n

2(n – r + 1) (r + 1) = (r + 1) r + (n – r + 1) (n – r)
\n\u21d2 2nr + 2n – 2r2<\/sup> – 2r + 2r + 2 = r2<\/sup> + r + n2<\/sup> – nr – nr + r2<\/sup> + n – r
\n\u21d2 2nr + 2n – 2r2<\/sup> + 2 = n2<\/sup> – 2nr + n + 2r2<\/sup>
\n\u21d2 n2<\/sup> – 2nr + n + 2r2<\/sup> – 2nr – 2n + 2r2<\/sup> – 2 = 0
\n\u21d2 n2<\/sup> – 4nr + 4r2<\/sup> – n – 2 = 0
\n\u21d2 n2<\/sup> – (4r + 1)n + 4r2<\/sup> – 2 = 0.<\/p>\n

\"TS<\/p>\n

Question 17.
\nIf \u2018P\u2019 and \u2018Q\u2019 are the sum of odd tenns and the sum of even terms respectively in the expansion of (x + a)n<\/sup> then prove that
\ni) P2<\/sup> – Q2<\/sup> = (x2<\/sup> – a2<\/sup>)n<\/sup>
\nii) 4PQ = (x + a)2n<\/sup> – (x – a)2n<\/sup> [AP – March 2016, March ’10]
\nSolution:
\nSince P and Q are the sum of odd terms and sum of even terms respectively in the expansion of (x + a)n<\/sup>.
\nNow (x + a)n<\/sup> =
\n\"TS<\/p>\n

i) L.HS: P2<\/sup> – Q2<\/sup>
\n= (P + Q) (P – Q) = (x + a) (x – a)
\n= [(x + a) (x – a)]n<\/sup>
\n= (x2<\/sup> – a2<\/sup>)n<\/sup> = R.H.S.<\/p>\n

ii) L.H.S : 4PQ = (P + Q)2<\/sup> – (P – Q)2<\/sup>
\n= ((x + a)n<\/sup>)2<\/sup> – ((x – a)n<\/sup>)2<\/sup>
\n= (x + a)2n<\/sup> – (x – a)2n<\/sup> = R.H.S.
\n\u2234 4PQ = (x + a)2n<\/sup> – (x – a)2n<\/sup>.<\/p>\n

Question 18.
\nIf the coefficients of 4 consecutive term in the expansion of (1 + x)n<\/sup> are a1<\/sub>, a2<\/sub>, a3<\/sub>, a4<\/sub> respectively then show that \\(\\frac{a_1}{a_1+a_2}+\\frac{a_3}{a_3+a_4}=\\frac{2 a_2}{a_2+a_3}\\). [AP – Mar. 2017; May\u201911. \u201807, March \u201911, ’95].
\nSolution:
\nSince a1<\/sub>, a2<\/sub>, a3<\/sub>, a4<\/sub>, are the coefficients of 4 consecutive terms in the expansion of (1 + x)n<\/sup>
\nLet a1<\/sub> = \\({ }^n C_{r-1}\\),
\na2<\/sub> = \\({ }^n C_r\\),
\na3<\/sub> = \\({ }^n C_{r+1}\\),
\na4<\/sub> = \\({ }^n C_{r+2}\\)
\nL.H.S:<\/p>\n

\"TS<\/p>\n

R.H.S:<\/p>\n

\"TS<\/p>\n

Question 19.
\nShow that the middle term in the expansion of (1 + x)2n<\/sup> is \\(\\frac{1 \\cdot 3 \\cdot 5 \\ldots \\ldots(2 n-1)}{n !}(2 x)^n\\).
\nSolution:
\nThe expansion of (1 + x)2n<\/sup> contains 2n + 1 terms.
\n\u2234 Middle term is (n + 1)th term
\n\u2234 Tn+1<\/sub> = \\({ }^{2 n} C_n\\) xn<\/sup><\/p>\n

= \\(\\frac{2 n !}{n ! \\cdot n !}\\) xn<\/sup><\/p>\n

= \\(\\frac{2 n(2 n-1)(2 n-2)(2 n-3) \\ldots . .4 \\times 3 \\times 2 \\times 1}{n ! n !} \\cdot x^n\\)<\/p>\n

= \\(\\frac{1 \\cdot 3 \\cdot 5 \\ldots .(2 n-1) \\cdot 2 \\times 4 \\times 6 \\times \\ldots .2 n}{n ! n !} \\cdot x^n\\)<\/p>\n

= \\(\\frac{1 \\cdot 3 \\cdot 5 \\ldots .(2 n-1)(1 \\times 2 \\times 3 \\times \\ldots \\ldots n) 2^n}{n ! n !} \\cdot x^n\\)<\/p>\n

\u2234 Middle term = \\(\\frac{1 \\cdot 3 \\cdot 5 \\ldots \\ldots(2 n-1)}{n !}(2 x)^n\\).<\/p>\n

\"TS<\/p>\n

Question 20.
\nIf (1 + 3x – 2x2<\/sup>)10<\/sup> = a0<\/sub> + a1<\/sub>x + a2<\/sub>x2<\/sup> + ……………. + ax20<\/sup> then prove that
\ni) a0<\/sub> + a1<\/sub> + a2<\/sub> + ………….. + a20<\/sub> = 210<\/sup>
\nii) a0<\/sub> – a1<\/sub> + a2<\/sub> – a3<\/sub> + ……………. + a20<\/sub> = 410<\/sup>
\nSolution:
\nGiven
\n(1 + 3x – 2x2<\/sup>)10<\/sup> = a0<\/sub> + a1<\/sub>x + a2<\/sub>x2<\/sup> + ……………. + ax20<\/sup> …………….(1)<\/p>\n

i) Put x = 1, in (1), we get
\n(1 + 3 – 2)10<\/sup> = a0<\/sub> + a1<\/sub> + a2<\/sub> + ………….. + a20<\/sub>
\n\u21d2 a0<\/sub> + a1<\/sub> + a2<\/sub> + …………….. + a20<\/sub> = 210<\/sup>.<\/p>\n

ii) Put x = – 1 in (1), we get
\n(1 – 3 – 2)\u2019\u00b0= a0<\/sub> – a1<\/sub> + a2<\/sub> – a3<\/sub> + ………………. + a20<\/sub>
\n\u21d2 a0<\/sub> – a1<\/sub> + a2<\/sub> – a3<\/sub> + ……………. + a20<\/sub> = 410<\/sup>.<\/p>\n

Question 21.
\nIf n is a positive integer, prove that \\(\\sum_{r=1}^n r^3 \\cdot\\left(\\frac{{ }^n C_r}{{ }^n C_{r-1}}\\right)^2=\\frac{n(n+1)^2(n+2)}{12}\\). [March ’13]
\nSolution:
\nNow, \\(\\frac{{ }^{n^n} C_r}{{ }^{n_C} C_{r-1}}=\\frac{\\frac{n !}{(n-r) ! r !}}{\\frac{n !}{(n-r+1) !(r-1) !}}\\)
\n= \\(\\frac{(n-r+1) !(r-1) !}{(n-r) ! r !}\\)<\/p>\n

\"TS<\/p>\n

Question 22.
\nFind the sum of the infinite series \\(1+\\frac{2}{3} \\cdot \\frac{1}{2}+\\frac{2 \\cdot 5}{3 \\cdot 6}\\left(\\frac{1}{2}\\right)^2+\\frac{2 \\cdot 5 \\cdot 8}{3 \\cdot 6 \\cdot 9}\\left(\\frac{1}{2}\\right)^3+\\ldots \\infty\\). [May ’91]
\nSolution:
\nLet the given series \u00a1s
\nS = 1 + \\(\\frac{2}{3} \\cdot \\frac{1}{2}+\\frac{2 \\cdot 5}{3 \\cdot 6} \\cdot \\frac{1}{2^2}+\\frac{2 \\cdot 5 \\cdot 8}{3 \\cdot 6 \\cdot 9} \\cdot \\frac{1}{2^3}+\\cdots \\cdots\\)
\nComparing with 1 + nx + \\(\\frac{n(n-1)}{1 \\cdot 2}\\) x2<\/sup> + …………….<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 23.
\n\\(\\frac{3 \\cdot 5}{5 \\cdot 10}+\\frac{3 \\cdot 5 \\cdot 7}{5 \\cdot 10 \\cdot 15}+\\frac{3 \\cdot 5 \\cdot 7 \\cdot 9}{5 \\cdot 10 \\cdot 15 \\cdot 20}+\\cdots \\cdots \\cdots \\infty\\) [TS – Mar. 2017; May ’09]
\nSolution:
\nLet the given series is
\nS = \\(\\frac{3 \\cdot 5}{5 \\cdot 10}+\\frac{3 \\cdot 5 \\cdot 7}{5 \\cdot 10 \\cdot 15}+\\frac{3 \\cdot 5 \\cdot 7 \\cdot 9}{5 \\cdot 10 \\cdot 15 \\cdot 20}+\\cdots \\cdots \\cdots \\infty\\)<\/p>\n

S + \\(\\frac{3}{5}\\) = \\(\\frac{3}{5}+\\frac{3 \\cdot 5}{5 \\cdot 10}+\\frac{3 \\cdot 5 \\cdot 7}{5 \\cdot 10 \\cdot 15}+\\frac{3 \\cdot 5 \\cdot 7 \\cdot 9}{5 \\cdot 10 \\cdot 15 \\cdot 20}+\\ldots \\ldots\\)<\/p>\n

\"TS<\/p>\n

Question 24.
\nIf x = \\(\\frac{1}{5}+\\frac{1 \\cdot 3}{5 \\cdot 10}+\\frac{1 \\cdot 3 \\cdot 5}{5 \\cdot 10 \\cdot 15}+\\ldots \\ldots \\cdots \\infty\\), then find 3x2<\/sup> + 6x. [TS – Mar. 2016, May ’14, ’11, ’08, Mar. ’14, ’12, ’06, TS – Mar. ’19]
\nSolution:
\nGiven x = \\(\\frac{1}{5}+\\frac{1 \\cdot 3}{5 \\cdot 10}+\\frac{1 \\cdot 3 \\cdot 5}{5 \\cdot 10 \\cdot 15}+\\ldots \\ldots \\ldots\\),
\nx + 1 = 1 + \\(\\frac{1}{5}+\\frac{1 \\cdot 3}{5 \\cdot 10}+\\frac{1 \\cdot 3 \\cdot 5}{5 \\cdot 10 \\cdot 15}+\\cdots \\cdots\\)
\nNow, comparing with<\/p>\n

\"TS<\/p>\n

\u21d2 x2<\/sup> + 1 + 2x = \\(\\frac{5}{3}\\)
\n\u21d2 3x2<\/sup> + 6x + 3 = 5
\n\u21d2 3x2<\/sup> + 6x = 2.<\/p>\n

Question 25.
\nFind the sum of the infinite series \\(\\frac{3}{4}+\\frac{3 \\cdot 5}{4 \\cdot 8}+\\frac{3 \\cdot 5 \\cdot 7}{4 \\cdot 8 \\cdot 12}+\\) [May ’10, March 11]
\nSolution:
\nLet the given series is
\nS = \\(\\frac{3}{4}+\\frac{3 \\cdot 5}{4 \\cdot 8}+\\frac{3 \\cdot 5 \\cdot 7}{4 \\cdot 8 \\cdot 12}+\\ldots \\ldots \\ldots\\)
\nS + 1 = 1 + \\(\\frac{3}{4}+\\frac{3 \\cdot 5}{4 \\cdot 8}+\\frac{3 \\cdot 5 \\cdot 7}{4 \\cdot 8 \\cdot 12}+\\ldots \\ldots \\ldots\\)
\nNow, comparing with<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 26.
\nIf x = \\(\\frac{1 \\cdot 3}{3 \\cdot 6}+\\frac{1 \\cdot 3 \\cdot 5}{3 \\cdot 6 \\cdot 9}+\\frac{1 \\cdot 3 \\cdot 5 \\cdot 7}{3 \\cdot 6 \\cdot 9 \\cdot 12}+\\ldots\\) then prove that 9x2<\/sup> + 24x = 11. [AP – MAr. ’19, ’17, ’15, May ’16; TS – MAr. ’18, ’16, Mar. ’09, Board Paper, May ’15]
\nSolution:
\nGiven<\/p>\n

\"TS<\/p>\n

Question 27.
\nIf x = \\(\\frac{5}{(2 !) \\cdot 3}+\\frac{5 \\cdot 7}{(3 !) \\cdot 3^2}+\\frac{5 \\cdot 7 \\cdot 9}{(4 !) \\cdot 3^3}+\\ldots\\) then find the value of x2<\/sup> + 4x. [TS – May; March ’13, May ’12]
\nSolution:
\nGiven,<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 28.
\nFind the sum to infinite terms of the series \\(\\frac{7}{5}\\left(1+\\frac{1}{10^2}+\\frac{1 \\cdot 3}{1 \\cdot 2} \\cdot \\frac{1}{10^4}+\\frac{1 \\cdot 3 \\cdot 5}{1 \\cdot 2 \\cdot 3} \\cdot \\frac{1}{10^6}+\\ldots\\right)\\). [AP – Mar. ’18, ’16; May ’13, March ’05].
\nSolution:<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 29.
\nShow that for any non-zero rational number x.
\n1 + \\(\\frac{x}{2}+\\frac{x(x-1)}{2 \\cdot 4}+\\frac{x(x-1)(x-2)}{2 \\cdot 4 \\cdot 6}+\\ldots \\ldots\\) = 1 + \\(\\frac{x}{3}+\\frac{x(x+1)}{3 \\cdot 6}+\\frac{x(x+1)(x+2)}{3 \\cdot 6 \\cdot 9}+\\ldots \\ldots\\) [March ’94]
\nSolution:
\nL.H.S = 1 + \\(\\frac{\\mathrm{n}}{2}+\\frac{\\mathrm{n}(\\mathrm{n}-1)}{2 \\cdot 4}+\\frac{\\mathrm{n}(\\mathrm{n}-1)(\\mathrm{n}-2)}{2 \\cdot 4 \\cdot 6}+\\ldots\\)<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Some More Maths 2A Binomial Theorem Important Questions<\/h3>\n

Question 1.
\nFind the 7th term in the expansion of \\(\\left(\\frac{4}{x^3}+\\frac{x^2}{2}\\right)^{14}\\).
\nSolution:
\nGiven \\(\\left(\\frac{4}{x^3}+\\frac{x^2}{2}\\right)^{14}\\)
\nHere, x = \\(\\frac{4}{x^3}\\); a = \\(\\frac{x^2}{2}\\); n = 14
\nr + 1 = 7
\n\u21d2 r = 6
\nThe general term in the expansion of (x + a)n<\/sup> is
\nTr+1<\/sub> = \\({ }^n C_r\\) xn-r<\/sup> ar<\/sup>
\nThe 7th term in the expansion of \\(\\left(\\frac{4}{x^3}+\\frac{x^2}{2}\\right)^{14}\\)<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the 5th term in the expansion of (3x – 4y)7<\/sup>.
\nSolution:
\nGiven (3x – 4y)7<\/sup>
\nHere, x = 3x; a = – 4y; n = 7
\nr + 1 = 5
\nr =4
\nThe general term in the expansion of (x + a)n<\/sup> is
\nTr + 1<\/sub> = \\({ }^n C_r\\) xn-r<\/sup> ar<\/sup>
\nThe rth term in the expansion of (3x – 4y)7<\/sup> is
\nT4+1<\/sub> = \\({ }^7 \\mathrm{C}_4\\) (3x)7-4<\/sup> (- 4y)4<\/sup>
\nT5<\/sub> = \\({ }^7 \\mathrm{C}_4\\) (3x)3<\/sup> (4y)4<\/sup>
\n= 35 . 33<\/sup> . x3<\/sup> 44<\/sup> . y4<\/sup>
\n= 241920 x3<\/sup> . y4<\/sup><\/p>\n

Question 3.
\nFind the middle term(s) in \\(\\left(\\frac{3}{a^3}+5 a^4\\right)^{20}\\).
\nSolution:
\nGiven \\(\\left(\\frac{3}{a^3}+5 a^4\\right)^{20}\\)
\nHere, x = \\(\\frac{3}{a^3}\\), a = 5a4<\/sup>, n = 20
\nSince, n = 20 is even,
\nMiddle term = \\(\\frac{n}{2}\\) + 1
\n= \\(\\frac{20}{2}\\) + 1
\n= 10 + 1 = 11th term
\nr + 1 = 11
\n\u21d2 r = 10
\nThe general term in this expansion is
\nTr+1<\/sub> = \\({ }^{{ }^n} C_r\\) xn-r<\/sup> ar<\/sup>
\n11th term of given expansion is
\nT10+1<\/sub> = \\({ }^{20} C_{10}\\left(\\frac{3}{a^3}\\right)^{20-10}\\left(5 a^4\\right)^{10}\\)
\n= \\({ }^{20} \\mathrm{c}_{10}\\left(\\frac{3}{a^3}\\right)^{10}\\left(5 a^4\\right)^{10}\\)<\/p>\n

T11<\/sub> = \\({ }^{20} C_{10}\\) . a-30<\/sup> . 510<\/sup> . a40<\/sup>
\n= \\({ }^{20} C_{10}\\) . 310<\/sup> . 510<\/sup> . a10<\/sup><\/p>\n

\"TS<\/p>\n

Question 4.
\nFind the middle term(s) in the expansion of (4x2<\/sup> + 5x3<\/sup>)17<\/sup>.
\nSolution:
\nGiven (4x2<\/sup> + 5x3<\/sup>)17<\/sup>
\nHere, x = 4x2<\/sup>, a = 5x3<\/sup>, n = 17
\nSince, n = 17 is odd, then
\nMiddle terms \\(\\frac{\\mathrm{n}+1}{2}, \\frac{\\mathrm{n}+3}{2}\\) = 9, 10 terms
\n9\u2019 term:
\nr + 1 = 9
\n\u21d2 r = 8
\nThe general term in this expansion is
\nTr+1<\/sub> = \\({ }^{{ }^n} C_r\\) xn-r<\/sup> ar<\/sup>
\n9th term of given expansion is
\nT8+1<\/sub> = \\({ }^{17} \\mathrm{c}_9\\) (4x2<\/sup>)17-8<\/sup> (5x3<\/sup>)8<\/sup>
\n= \\({ }^{17} \\mathrm{c}_9\\) (4x2<\/sup>)9<\/sup> (5x3<\/sup>)8<\/sup>
\n= \\({ }^{17} \\mathrm{c}_9\\) . 49<\/sup> . x18<\/sup> . 58<\/sup> . x24<\/sup>
\n= \\({ }^{17} \\mathrm{c}_9\\) . 49<\/sup> . 58<\/sup> . x42<\/sup><\/p>\n

10th term:
\nr + 1 = 10
\n\u21d2 r = 9
\nThe general term in this expansion is
\nTr+1<\/sub> = \\({ }^{{ }^n} C_r\\) xn-r<\/sup> ar<\/sup>
\n10th term in this expansion \u00a1s
\nT9+1<\/sub> = \\({ }^{17} \\mathrm{C}_9\\)(4x2<\/sup>)17-9<\/sup> (5x3<\/sup>)9<\/sup>
\nT10<\/sub> = \\({ }^{17} \\mathrm{C}_9\\) (4x2<\/sup>)8<\/sup> (5x3<\/sup>)9<\/sup>
\n= \\({ }^{17} \\mathrm{C}_9\\) . 48<\/sup> . 59<\/sup> . x16<\/sup> . x27<\/sup>
\n= \\({ }^{17} \\mathrm{C}_9\\) . 48<\/sup> . 59<\/sup> . x43<\/sup><\/p>\n

Question 5.
\nFind the coefficient of x11<\/sup> in \\(\\left(2 x^2+\\frac{3}{x^3}\\right)^{13}\\).
\nSolution:
\nGiven \\(\\left(2 x^2+\\frac{3}{x^3}\\right)^{13}\\)
\nHere, x = 2x2<\/sup>,
\na = \\(\\frac{3}{x^3}\\), n = 13
\nNow, the general term in this expansion is
\nTr+1<\/sub> = \\(\\) xn-r<\/sup> ar<\/sup>
\n= \\({ }^{13} \\mathrm{C}_{\\mathrm{r}}\\) (2xr<\/sup>)13-r<\/sup> \\(\\left(\\frac{3}{x^3}\\right)^r\\)
\n= \\({ }^{13} \\mathrm{C}_{\\mathrm{r}}\\) 213-r<\/sup> x26-2r<\/sup> . 32<\/sup> . x-3r<\/sup>
\n= \\({ }^{13} \\mathrm{C}_{\\mathrm{r}}\\) 213-r<\/sup> 3r<\/sup> x26-5r<\/sup> ………………(1)
\nTo find the coefficient of x11<\/sup>
\nPut 26 – 5r = 11
\n5r = 15
\n\u21d2 r = 3
\nNow substituting r = 3 in equation (1) we get
\nT3+1<\/sub> = \\({ }^{13} \\mathrm{C}_3\\) 213-3<\/sup> 33<\/sup> x26-15<\/sup>
\nT4<\/sub> = \\({ }^{13} \\mathrm{C}_3\\) 210<\/sup> 33<\/sup> x11<\/sup>
\nThe coefficient of x11<\/sup> in the expansion of \\(\\left(2 x^2+\\frac{3}{x^3}\\right)^{13}\\) is \\({ }^{13} \\mathrm{C}_3\\) 210<\/sup> 33<\/sup>.<\/p>\n

Question 6.
\nFind the term independent of x in the expansion of \\(\\left(\\sqrt{\\frac{x}{3}}+\\frac{3}{2 x^2}\\right)^{10}\\).
\nSolution:
\nGiven \\(\\left(\\sqrt{\\frac{x}{3}}+\\frac{3}{2 x^2}\\right)^{10}\\)
\nHere, x = \\(\\sqrt{\\frac{x}{3}}\\), a = \\(\\frac{3}{2 x^2}\\), n = 10
\nThe general term in this expansion is
\nTr+1<\/sub> = \\({ }^n C_r\\) xn-r<\/sup> ar<\/sup>
\nTr+1<\/sub> = \\({ }^{10} C_r\\left(\\sqrt{\\frac{x}{3}}\\right)^{10-r}\\left(\\frac{3}{2 x^2}\\right)^r\\)
\n= \\({ }^{10} C_r \\frac{x^{\\frac{10-r}{2}}}{3^{\\frac{10-r}{2}}} \\cdot \\frac{3^r}{2^r} \\cdot x^{-2 r}\\)
\n= \\({ }^{10} \\mathrm{C}_{\\mathrm{r}} \\frac{3^{\\mathrm{r}}}{3^{\\frac{10-\\mathrm{r}}{2}} \\cdot 2^{\\mathrm{r}}} \\mathrm{x}^{\\frac{10-\\mathrm{r}}{2}-2 \\mathrm{r}}\\) ……………….(1)
\nTo find the term independent of x11<\/sup> (i.e., coeff. of x\u00b0)
\nPut \\(\\frac{10-\\mathrm{r}}{2}\\) – 2r = 0
\n10 – r – 4r = 0
\n5r = 10
\n\u21d2 r = 2
\nSubstitute r = 2 in equation (1) we get
\nT = \\({ }^{10} \\mathrm{C}_2 \\frac{3^2}{3^{\\frac{10-2}{2}} \\cdot 2^2} \\cdot \\mathrm{x}^{\\frac{10-2}{2}-2^2}\\)
\n= \\({ }^{10} C_2 \\frac{3^2}{3^3 \\cdot 2^2} \\cdot x^0\\)
\n= \\(\\frac{10 \\cdot 9}{1 \\cdot 2} \\times \\frac{3^2}{3^4 \\cdot 2^2}=\\frac{10 \\cdot 9}{1 \\cdot 2} \\cdot \\frac{1}{9 \\cdot 4} \\cdot x^0=\\frac{5}{4} x^0\\)
\nThe term independent of x in the given expansion is T3<\/sub> = \\(\\frac{5}{4}\\).<\/p>\n

\"TS<\/p>\n

Question 7.
\nFind the largest binomial coefficients In the expansion of (1 + x)19<\/sup>.
\nSolution:
\nGiven, (1 + x)19<\/sup>
\nHere n = 19, an odd integer.
\nThe largest binomial coefficients are \\({ }^n C_{\\frac{n-1}{2}}\\) and \\({ }^n C_{\\frac{n+1}{2}}\\)
\n\\({ }^{19} \\mathrm{C}_{\\frac{19-1}{2}} \\text { and }{ }^{19} \\mathrm{C}_{\\frac{19+1}{2}}\\) = \\({ }^{19} C_9 \\text { and }{ }^{19} C_{10}\\)
\n(Note that, \\({ }^{19} C_{9}={ }^{19} C_{10}\\)).<\/p>\n

Question 8.
\nIf the coefficients of (2r + 4)th, (r – 2)th terms in the expansion of (1 + x)21<\/sup> equal find \u2018r\u2019.
\nSolution:
\nGiven (1 – x)18<\/sup>
\nThe general term in the expansion of (x + a)n<\/sup> is
\nTr+1<\/sub> = \\({ }^n C_r\\) xn-r<\/sup> ar<\/sup>
\n(2r + 4)th term in the expansion of (1 + x)18<\/sup> is
\nT(2r+3)+1<\/sub> = \\({ }^{18} \\mathrm{C}_{2 \\mathrm{r}+3}\\) (1) x2r+3<\/sup>
\nT2r+4<\/sub> = \\({ }^{18} \\mathrm{C}_{2 \\mathrm{r}+3}\\) x2r+3<\/sup>
\nThe coeff. of (2r + 4)th term is \\({ }^{18} \\mathrm{C}_{2 \\mathrm{r}+3}\\)
\n(r – 2) term in the expansion of (1 + x)18<\/sup> is
\nT(r-3)+1<\/sub> = \\({ }^{18} \\mathrm{C}_{\\mathrm{r}-3}\\) 118-(r-3)<\/sup> xr-3<\/sup>
\nTr-2<\/sub> = \\({ }^{18} \\mathrm{C}_{\\mathrm{r}-3}\\) . xr-3<\/sup>
\nThe coeff. of (r – 2)nd term is \\({ }^{18} \\mathrm{C}_{\\mathrm{r}-3}\\)
\n\u2234 Given that two coefficients are equal.
\n\\({ }^{18} \\mathrm{C}_{2 \\mathrm{r}+3}={ }^{18} \\mathrm{C}_{\\mathrm{r}-3}\\)
\n\\({ }^n C_r={ }^n C_s\\)
\n\u21d2 n = r + s (or) r = s
\n18 = 2r + 3 + r – 3
\n2r + 3 = r – 3
\n3r = 18
\n\u21d2 r = 6
\n\u21d2 r = – 6
\nSince, r is a positive integer, we get r = 6.<\/p>\n

Question 9.
\nFind the set E of x for which the binomial expansion (7 + 3x)-5<\/sup>.
\nSolution:
\nGiven (7 + 3x)-5<\/sup>
\n7-5<\/sup> (1 + \\(\\frac{3 x}{7}\\))-5<\/sup>
\nThe binomial expansion of (7 + 3x)-5<\/sup> is valid when
\n\\(\\left|\\frac{3 x}{7}\\right|<1 \\Rightarrow|x|<\\frac{7}{3}\\)
\nx \u2208 \\(\\left(\\frac{-7}{3}, \\frac{7}{3}\\right)\\)<\/p>\n

Question 10.
\nFind the set E of x for which the binomial expansion (7 – 4x)-5<\/sup> is valid.
\nSolution:
\nGiven (7 – 4x)-5<\/sup> = 7-5<\/sup> (1 – \\(\\frac{4 x}{7}\\))-5<\/sup>
\nThe binomial expansion of (7 – 4x)-5<\/sup> is valid when
\n\\(\\left|\\frac{-4 x}{7}\\right|<1 \\Rightarrow\\left|\\frac{4 x}{7}\\right|<1\\)
\n\u21d2 x < \\(\\frac{7}{4}\\)
\n\u21d2 x \u2208 \\(\\left(\\frac{-7}{4}, \\frac{7}{4}\\right)\\)
\n\u2234 E = \\(\\left(\\frac{-7}{4}, \\frac{7}{4}\\right)\\)<\/p>\n

\"TS<\/p>\n

Question 11.
\nFind the 6th term of (1 + \\(\\frac{x}{2}\\))-5<\/sup>
\nSolution:
\nGiven, (1 + \\(\\frac{x}{2}\\)))-5<\/sup>
\nComparing this with (1 + x)n<\/sup>, where
\nx = \\(\\frac{x}{2}\\), n = – 5,
\n\u21d2 r + 1 = 6
\n\u21d2 r = 5
\nThe general term in the binomial expansion of (1 + x)n<\/sup> is<\/p>\n

\"TS<\/p>\n

Question 12.
\nFind the coefficient of x6<\/sup> in (3x – \\(\\frac{4}{x}\\))10<\/sup>.
\nSolution:
\nGiven (3x – \\(\\frac{4}{x}\\))10<\/sup>
\nHere, x = 3x; a = – \\(\\frac{4}{x}\\), n = 10
\nThe general term in this expansion is
\nTr+1<\/sub> = \\({ }^n C_r\\) xn-r<\/sup> ar<\/sup>
\n= \\({ }^{10} \\mathrm{C}_{\\mathrm{r}}\\) (3x)10-r<\/sup> \\(\\left(\\frac{-4}{x}\\right)^{\\mathbf{r}}\\)
\n= \\({ }^{10} \\mathrm{C}_{\\mathrm{r}}\\) 310-r<\/sup> x10-r<\/sup> (- 4)r<\/sup> xr<\/sup>
\n= \\({ }^{10} \\mathrm{C}_{\\mathrm{r}}\\) 310-r<\/sup> . (- 4)10-r<\/sup> . x10-2r<\/sup> …………..(1)
\nTo find the coefficient of x-6<\/sup>,
\nput 10 – 2r = – 6
\n\u21d2 2r = 16
\n\u21d2 r = 8
\nNow, substituting r = 8 in equation (1) we get
\nT8+1<\/sub> = \\({ }^{10} \\mathrm{C}_8\\) 310-8<\/sup> (- 4)8<\/sup> x10-16<\/sup>
\nT9<\/sub> = \\({ }^{10} \\mathrm{C}_8\\) 32<\/sup> (4)8<\/sup> x-6<\/sup>
\nThe coeff. of x-6<\/sup> in the expansion of (3x – \\(\\frac{4}{x}\\))10<\/sup> is \\({ }^{10} \\mathrm{C}_8\\) . 32<\/sup> . (4)8<\/sup><\/p>\n

Question 13.
\nFind the middle term (s) in the expansion of \\(\\left(4 a+\\frac{3}{2} b\\right)^{11}\\).
\nSolution:
\nGiven \\(\\left(4 a+\\frac{3}{2} b\\right)^{11}\\)
\nHere, x = 4a; a = \\(\\frac{3}{2}\\) b; n = 11
\nSince, n = 11 is odd,
\nMiddle terms = \\(\\frac{\\mathrm{n}+1}{2}, \\frac{\\mathrm{n}+3}{2}\\) = 6, 7 terms.<\/p>\n

6th term :
\nr + 1 = 6
\n\u21d2 r = 5
\nThe general term in this expansion is
\nTr+1<\/sub> = \\({ }^n C_r\\) xn-r<\/sup> ar<\/sup>
\n6th term of given expansion \u00a1s
\nT5+1<\/sub> = \\({ }^{11} C_5(4 a)^{11-5}\\left(\\frac{3}{2} b\\right)^5\\)
\nT6<\/sub> = \\({ }^{11} \\mathrm{C}_5(4 \\mathrm{a})^6\\left(\\frac{3}{2} \\mathrm{~b}\\right)^5\\)
\n= \\(=11 C_5 \\frac{4^6 \\cdot 3^5}{2^5} \\cdot a^6 \\cdot b^5\\)
\nT6<\/sub> = \\({ }^{11} \\mathrm{C}_6 \\frac{2^{12} \\cdot 3^5}{2^5}\\) . a4<\/sup>b5<\/sup>
\n= \\({ }^{11} \\mathrm{C}_6\\) . a6<\/sup> . b5<\/sup><\/p>\n

7th term :
\nr + 1 = 7
\n\u21d2 r = 6
\nThe general term in this expansion is
\nTr+1<\/sub> = \\({ }^n C_r\\) xn-r<\/sup> ar<\/sup>
\n7th term of given expansion is<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 14.
\nFind the middle term (s) in the expansion of (4x2<\/sup> + 5x3<\/sup>)17<\/sup>.
\nSolution:
\nGiven (4x2<\/sup> + 5x3<\/sup>)17<\/sup>.
\nHere, x = 4x2<\/sup>, a = 5x3<\/sup>, n= 17.
\nSince, n = 17 is odd, then
\nmiddle terms = \\(\\frac{n+1}{2}, \\frac{n+3}{2}\\) = 9, 10 terms<\/p>\n

9th term :
\nr + 1 = 9
\n\u21d2 r = 8
\nThe general term in this expansion is
\nTr+1<\/sub> = \\({ }^n C_r\\) xn-r<\/sup> ar<\/sup>
\n\u2234 9th term in this expansion is
\nT9+1<\/sub> = \\({ }^{17} \\mathrm{C}_9\\)(4x2<\/sup>)17-9<\/sup> (5x3<\/sup>)9<\/sup>
\nT10<\/sub> = \\({ }^{17} \\mathrm{C}_9\\) (4x2<\/sup>)8<\/sup> (5x3<\/sup>)9<\/sup>
\n= \\({ }^{17} \\mathrm{C}_9\\) 48<\/sup> . 59<\/sup> x16<\/sup> . x27<\/sup>
\n= \\({ }^{17} \\mathrm{C}_9\\) 48<\/sup> . 59<\/sup> . x43<\/sup><\/p>\n

Question 15.
\nProve that 2 . C0<\/sub> + 5 . C1<\/sub> + 8 . C2<\/sub> + ……………… + (3n + 2) Cn<\/sub> = (3n + 4) 2n-1<\/sup>
\nSolution:
\nLet S = 2 . C0<\/sub> + 5 . C1<\/sub> + 8 . C2<\/sub> + ……………… + (3n + 2) Cn<\/sub> = (3n + 4) 2n-1<\/sup> …………….(1)
\nBy writing the terms in (1) in the reverse order
\nS = (3n + 2) . Cn<\/sub> + (3n – 1) . Cn-1<\/sub> + (3n – 4) Cn-2<\/sub> + ……………….. + 2. C0<\/sub>
\nS = (3n + 2) C0<\/sub> + (3n – 1) C1<\/sub> + (3n – 4) C2<\/sub> + ……………… + 2 Cn<\/sub> ………….. (2)
\nAdding (1) and (2) we get,
\nS = 2 . C0<\/sub> + 5 . C1<\/sub> + 8 . C2<\/sub> + …………… + (3n + 2) . Cn<\/sub>
\nS = (3n + 2) . C0<\/sub> + (3n – 1) . C1<\/sub> + (3n – 4) . C2<\/sub> + ……………. + 2 . Cn<\/sub>
\n2S = (3n + 4)C0<\/sub> + (3n + 4) C1<\/sub> . (3n + 4) C2<\/sub> + …………… + (3n + 4) Cn<\/sub>
\n2S = (3n + 4) (C0<\/sub> + C1<\/sub> + C2<\/sub> + …………….. + Cn<\/sub>)
\n2S = (3n + 4) 2n<\/sup>
\n\u21d2 S = (3n + 4) 2n-1<\/sup>
\n\u2234 2 . C0<\/sub> + 5 . C1<\/sub> + 8. C2<\/sub> + …………… + (3n + 2) Cn<\/sub> = (3n + 4) . 2n-1<\/sup><\/p>\n

Question 16.
\nProve that
\n(C0<\/sub> + C1<\/sub>) (C1<\/sub> + C1<\/sub>) (C2<\/sub> + C3<\/sub>) + ………………… + (Cn-1<\/sub> + Cn<\/sub>) = \\(\\frac{(n+1)^n}{n !}\\) . C0<\/sub> . C1<\/sub> . C2<\/sub> ……………. Cn<\/sub><\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 17.
\nWrite the first 3 terms in the expansion of \\(\\left(1+\\frac{\\mathrm{x}}{2}\\right)^{-5}\\).
\nSolution:
\nGiven \\(\\left(1+\\frac{\\mathrm{x}}{2}\\right)^{-5}\\)
\nWe know that,<\/p>\n

\"TS<\/p>\n

Hence, the first three terms in the expansion of \\(\\left(1+\\frac{\\mathrm{x}}{2}\\right)^{-5}\\) are 1, \\(\\frac{-5 x}{2}, \\frac{15 x^2}{4}\\).<\/p>\n

Question 18.
\nFind the coefficient of x6<\/sup> in the expansion of (1 – 3x)-2\/5<\/sup>.
\nSolution:
\nGiven that (1 + 3x)-2\/5<\/sup>
\nThe general term of (1 – 3x)-2\/5<\/sup> is
\nTr+1<\/sub> = \\(\\frac{n(n-1)(n-2) \\ldots .(n-r+1)}{r !}\\)
\nwhere x = – 3x, n = \\(\\frac{-2}{5}\\)<\/p>\n

Tr+1<\/sub> = \\(\\frac{\\frac{-2}{5}\\left(\\frac{-2}{5}-1\\right)\\left(\\frac{-2}{5}-2\\right) \\ldots \\ldots . .\\left(\\frac{-2}{5}-r+1\\right)}{r !}(-3 x)^r\\)<\/p>\n

= \\(\\frac{\\frac{-2}{5}\\left(\\frac{-7}{5}\\right)\\left(\\frac{-12}{5}\\right) \\cdots \\cdots \\cdots\\left(\\frac{3}{5}-\\mathrm{r}\\right)}{\\mathrm{r} !}(-3 \\mathrm{r})^{\\mathrm{r}}\\)<\/p>\n

= \\(\\frac{2 \\cdot 7 \\cdot 12 \\ldots \\ldots(5 \\mathrm{r}-3)}{5^{\\mathrm{r}} \\cdot \\mathrm{r} !}(3 \\mathrm{x})^{\\mathrm{r}}\\)<\/p>\n

Put r = 6
\nThen the coefficient of x-6<\/sup> is \\(\\frac{2 \\cdot 7 \\cdot 12 \\ldots \\ldots .27}{5^6 \\cdot 6 !}(3)^6\\)
\n= \\(\\frac{2 \\cdot 7 \\cdot 12 \\ldots \\ldots .27}{6 !} \\cdot\\left(\\frac{3}{5}\\right)^6\\).<\/p>\n

Question 19.
\nEind the coefficient of x4<\/sup> in the expansion of (1 – 4x)3\/5<\/sup>
\nSolution:
\nGiven (1 – 4x)3\/5<\/sup>
\nThe general term of (1 – 4x)3\/5<\/sup> is
\nTr+1<\/sub> = \\(\\frac{n(n-1)(n-2) \\ldots \\ldots(n-r+1)}{r !}\\) . xr<\/sup>
\nwhere, x = – 4x, n = \\(\\frac{-3}{5}\\)<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 20.
\nFind the sum of the infinite series 1 + \\(\\frac{1}{3}+\\frac{1 \\cdot 3}{3 \\cdot 6}+\\frac{1 \\cdot 3 \\cdot 5}{3 \\cdot 6 \\cdot 9}+\\ldots \\ldots\\). [TS – Mar. 2015]
\nSolution:
\nLet the given series is
\nS = 1 + \\(\\frac{1}{3}+\\frac{1 \\cdot 3}{3 \\cdot 6}+\\frac{1 \\cdot 3 \\cdot 5}{3 \\cdot 6 \\cdot 9}+\\ldots \\ldots\\)
\nNow, comparing with<\/p>\n

\"TS<\/p>\n

Question 21.
\nFind the sum of the infinite series 1 – \\(\\frac{4}{5}+\\frac{4 \\cdot 7}{5 \\cdot 10}-\\frac{4 \\cdot 7 \\cdot 10}{5 \\cdot 10 \\cdot 15}\\) + …………..
\nSolution:
\nLet the given series is
\nS = 1 – \\(\\frac{4}{5}+\\frac{4 \\cdot 7}{5 \\cdot 10}-\\frac{4 \\cdot 7 \\cdot 10}{5 \\cdot 10 \\cdot 15}\\) + …………..
\nNow comparing with<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 22.
\nFind the sum of the infinite series \\(\\frac{3}{4 \\cdot 8}-\\frac{3 \\cdot 5}{4 \\cdot 8 \\cdot 12}+\\frac{3 \\cdot 5 \\cdot 7}{4 \\cdot 8 \\cdot 12 \\cdot 16}\\).
\nSolution:
\nLet the given series<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 23.
\nIf t = \\(\\frac{4}{5}+\\frac{4 \\cdot 6}{5 \\cdot 10}+\\frac{4 \\cdot 6 \\cdot 8}{5 \\cdot 10 \\cdot 15}\\) + ……………….. \u221e, then prove that 9t = 16.
\nSolution:
\nGiven<\/p>\n

\"TS<\/p>\n

Here p = 4; p + q = 6;
\n\u21d2 q = 2
\n\u21d2 x = \\(\\frac{2}{5}\\)
\n\u2234 (1) \u21d2 1 + t = \\(\\left(1-\\frac{2}{5}\\right)^{\\frac{-4}{2}}\\)
\n\u21d2 1 +t = \\(\\left(\\frac{3}{5}\\right)^{-2}\\)
\n\u21d2 1 + t = \\(\\frac{25}{9}\\)
\n\u21d2 9t =16.<\/p>\n

\"TS<\/p>\n

Question 24.
\nIf I, n are positive integers, 0 < f < 1 and if (7 + 4\u221a3)n<\/sup> = 1 + f, then show that
\ni) I is an odd integer and
\nii) (I + f) (1 – f) = 1.
\nSolution:
\nSol. Since I, n are positive integers 0 < F < 1 and (7 + 4\u221a3)n<\/sup> = 1 + F
\nNow, 36 < 48 < 49
\n\u21d2 6 < 4\u221a3 < 7 \u21d2 – 6 > – 4\u221a3 > – 7
\n\u21d2 1 > 7 – 4\u221a3 > 0
\n\u21d2 0 < 7 – 4\u221a3 < 1
\n\u21d2 0< 7 – 4\u221a3 < 1
\n\u21d2 o < (7 – 4\u221a3)n<\/sup> < 1
\nLet (7 – 4\u221a3)n<\/sup> = f
\n\u2234 0 < f < 1
\nNow, 1 + f = (7 + 4\u221a3)n<\/sup><\/p>\n

\"TS<\/p>\n

where k is an integer
\n\u21d2 I + F + f = Even integer.
\nSince, I is an integer, F + f is an integer.
\nBut, 0 < F < 1, 0 < f < 1
\n\u21d2 0 < F + f < 2.
\n\u2234 f + F = 1
\nFrom (1)
\n\u21d2 I + 1 = Even integer
\n\u21d2 Even integer – 1 = Odd Integer
\n\u2234 I is an odd integer.<\/p>\n

ii) L.H.S: (I + f) (I – f) = (7 + 4\u221a3)n<\/sup> . F
\n= (7 + 4\u221a3)n<\/sup> (7 – 4\u221a3)n<\/sup>
\n= (49 – 48)n<\/sup> = 1n<\/sup> = 1 = R.H.S
\n\u2234 LH.S = R.H.S.<\/p>\n

Question 25.
\nFind the 8th term of \\(\\left(1-\\frac{5 x}{2}\\right)^{-3 \/ 5}\\). [AP – Mar. ’18]
\nSolution:
\nGiven \\(\\left(1-\\frac{5 x}{2}\\right)^{-3 \/ 5}\\)
\nT8<\/sub> = \\(\\frac{\\left(\\frac{-3}{5}\\right)\\left(\\frac{-3}{5}-1\\right)\\left(\\frac{-3}{5}-2\\right) \\ldots\\left(\\frac{-3}{5}-6\\right)}{7 !}\\left(\\frac{-5 x}{2}\\right)^7\\)
\n= \\(\\frac{\\left(\\frac{3}{5}\\right)\\left(\\frac{8}{5}\\right)\\left(\\frac{13}{5}\\right) \\ldots\\left(\\frac{33}{5}\\right)}{7 !}\\left(\\frac{5 x}{2}\\right)^7\\)
\n= \\(\\frac{3 \\times 8 \\times 13 \\times \\ldots . \\times 33}{7 !}\\left(\\frac{x}{2}\\right)^7\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

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