TS Inter Second Year Maths 2A Binomial Theorem Important Questions Long Answer Type

Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Binomial Theorem Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter Second Year Maths 2A Binomial Theorem Important Questions Long Answer Type

Question 1.
State and prove Binomial Theorem. [March ’09]
Solution:

Question 2.
Find the numerically greatest term in the expansion of (4 + 3x)¹⁵ where x = \(\frac{7}{2}\).
Solution:
Given
(4 + 3x)¹⁵
⇒ 4¹⁵ (1 + \(\frac{3 x}{4}\))¹⁵ = 4¹⁵ (1 + x)ⁿ
where n = 15; x = \(\frac{3 x}{4}\)
\(|x|=\left|\frac{3 x}{4}\right|=\left|\frac{3}{4} \cdot \frac{7}{2}\right|=\frac{21}{8}\)
Now, \(\frac{(n+1)|x|}{|x|+1}=\frac{(15+1)\left|\frac{3 x}{4}\right|}{\left|\frac{3 x}{4}\right|+1}\)
= \(\frac{16 \cdot \frac{21}{8}}{\frac{21}{8}+1}=\frac{336}{29}\) = 11.5
∴ T₁₂ is numerically greatest.
r + 1 = 12
⇒ r = 11
The general term in the expansion of (x + a)ⁿ is
T_{r + 1} = \({ }^n C_r\) x^{n – r} a^r
T_{11 + 1} = \({ }^{15} \mathrm{C}_{11}\) (4)^{15 – 11} (3x)¹¹
12th term in this expansion is

Question 3.
Find the numerically reatest tenu in the expansion of (4a – 6b)¹³ when a = 3, b = 5.
Solution:
Given (4a – 6b)¹³ = (4a)¹³ (1 – \(\frac{6 b}{4 a}\))¹³
= (4a)¹³ (1 + x)ⁿ
where, x = \(-\frac{6 b}{4 a}\), n = 13
|x| = \(\left|\frac{-6 b}{4 a}\right|=\left|\frac{-6 \cdot 5}{4 \cdot 3}\right|=\frac{5}{2}\)
Now, \(\frac{(n+1)|x|}{|x|+1}=\frac{(13+1) \frac{5}{2}}{\frac{5}{2}+1}\)
= \(\frac{14 \cdot 5}{5+2}=\frac{70}{7}\) = 10
∴ T₁₀ and T₁₁ are numerically greatest.
T₁₀:
r + 1 = 10
⇒ r = 9
The general term in the expansion of (1 + x)ⁿ is
T_{r + 1} = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) x^{n – r} a^r
T₁₀ in this expansion is
T₉₊₁ = \({ }^{13} \mathrm{C}_9\) (4a)^13-9 (- 6b)⁹
T10 = \({ }^{13} \mathrm{C}_9\) (4a)⁴ (- 6b)⁹
= \({ }^{13} \mathrm{C}_9\) (4⁴ . 3⁴) ((- 6)⁹ (5)⁹)
= \({ }^{13} \mathrm{C}_9\) 12⁴ . 30⁹
|T₁₀| = \({ }^{13} \mathrm{C}_9\) 12⁴ . 30⁹

T₁₁:
r + 1 = 11
⇒ r = 10.
The general term in the expansion of (1 – x)ⁿ is
T_{r + 1} = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) x^{n – r} a^r
T₁₁ in this expansion is
T₁₀₊₁ = \({ }^{13} \mathrm{C}_{10}\) (4a)^13-10 . (- 6b)¹⁰
= \({ }^{13} \mathrm{C}_{10}\) (4a)³ (6b)¹⁰
T₁₁ = \({ }^{13} \mathrm{C}_{10}\) (4 . 3)³ (6 . 5)¹⁰
= \({ }^{13} \mathrm{C}_{10}\) 12³ . 30¹⁰
T₁₁ = \({ }^{13} \mathrm{C}_{10}\) 12³ . 30¹⁰
∴ |T₁₀| = |T₁₁|.

Question 4.
If the coefficient of x¹⁰ in the expansion of \(\left(a x^2+\frac{1}{b x}\right)^{11}\) is equal to the coefficient of x^-10 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]
Solution:
Case – I :
Given \(\left(a x^2+\frac{1}{b x}\right)^{11}\)
Here x = ax², a = \(\frac{1}{\mathrm{bx}}\); n = 11
Now, the general term in the expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
= \({ }^{11} \mathrm{C}_{\mathrm{r}}\) a^11-r x^22-2r 1^r b^-r x^-r
= \({ }^{11} \mathrm{C}_{\mathrm{r}}\) a^11-r b^-r x^22-3r …………….(1)
To find the coefficient of x¹⁰,
Put 22 – 3r = 10
3r = 12
r = 4
Substituting r = 4 in equation (1) we get
T₄₊₁ = \({ }^{11} C_4\) a^11-4 b^-4 x^22-12
T₅ = \({ }^{11} C_4\) a⁷ b^-4 x¹⁰
T₅ = \({ }^{11} C_4\) a⁷ b^-4 x¹⁰
∴ The coeff. of x¹⁰ ¡n the expansion of \(\left(a x^2+\frac{1}{b x}\right)^{11}\) is \({ }^{11} C_4\) a⁷ b^-4

Case II:
Given \(\left(a x-\frac{1}{b x^2}\right)^{11}\)
Here x = ax, a = \(\left(\frac{-1}{b x^2}\right)\), n = 11
Now, the general term in the expansion is
T_{r + 1} = \({ }^n C_r\) x^n-r a^r
= \({ }^{11} C_r(a x)^{11-r}\left(\frac{-1}{b x^2}\right)^r\)
= \({ }^{11} \mathrm{C}_{\mathrm{r}}\) a^11-r x^11-r (- 1)^r b^-r x^-2r
= \({ }^{11} \mathrm{C}_{\mathrm{r}}\) (- 1)^r b^-r x^11-3r ……………(2)
To find the coefficient of x^-10
put 11 – 3r = – 10
⇒ 3r = 21
⇒ r = 7
Substitute r = 7 in equation (2) we get
T₇₊₁ = \({ }^{11} \mathrm{C}_7\) a^11-7 b⁷ (- 1)⁷ x^11-21
T₈ = – \({ }^{11} \mathrm{C}_7\) a⁴ b^-7 x^-10
∴ The coeff. of x^-10 in the expansion of \(\left(a x-\frac{1}{b x^2}\right)^{11}\) is – \({ }^{11} \mathrm{C}_7\) a⁴ b^-7
Given that, these coefficients are equal
\({ }^{11} \mathrm{C}_7\) a⁷ b^-4 = – \({ }^{11} \mathrm{C}_7\) a⁴ b^-7
\({ }^{11} C_4 \frac{a^7}{b^4}=-{ }^{11} C_4 \frac{a^4}{b^7}\)
\(\frac{a^7}{b^4}=\frac{-a^4}{b^7}\)
⇒ a³b³ = – 1
⇒ (ab)³ = (- 1)³
⇒ ab = – 1

Question 5.
If n is a positive integer, then show that
i) C₀ + C₁ + C₂ + ……………… + C_n = 2
ii) C₀ + C₂ + C₄ + ……………. = C₁ + C₃ + C₅ + ……….. = 2^n-1 [March ’97, ’90]
Solution:
We know that
(1 + x)ⁿ = C₀ + C₁x + C₂x² + ……………… + C_nxⁿ ………….(1)
i) Put x = 1 in equation (1)
⇒ (1 + 1)ⁿ = C₀ + C₁(1) + C₂(1)² + C₃(1)³ + C₄(1)⁴ + ……………… + C_n(1)ⁿ
2ⁿ = C₀ + C₁ + C₂ + C₃ + C₄ + …………. + C_n
∴ C₀ + C₁ + C₂ + C₃ + C₄ + …………. + C_n = 2ⁿ

ii) Put x = – 1 in equation (1), we get
(1 – 1)ⁿ = C₀ + C₁ (- 1) + C₂ (- 1)² + C₃ (- 1)³ + ……………. + C_n (- 1)ⁿ
o = C₀ – C₁ + C₂ – C₃ + C₄ – ………………
C₀ + C₂ + C₄ + …………………. = C₁ + C₃ + C₅ + ………….
Since a = b
⇒ a = b = \(\frac{a+b}{2}\)
C₀ + C₂ + C₄ + …………………. = C₁ + C₃ + C₅ + ………….
= \(\frac{\mathrm{C}_0+\mathrm{C}_2+\mathrm{C}_4+\ldots \ldots+\mathrm{C}_1+\mathrm{C}_3+\mathrm{C}_5+\ldots \ldots}{2}\)
= \(\frac{\mathrm{C}_0+\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3+\mathrm{C}_4+\mathrm{C}_5+\ldots \ldots}{2}\)
= \(\frac{2^{\mathrm{n}}}{2}\)
= 2^{n – 1}
∴ C₀ + C₂ + C₄ + …………………. = C₁ + C₃ + C₅ + …………. = 2^{n – 1}
C₀ + C₂ + C₄ + …………………. = 2^{n – 1}
C₁ + C₃ + C₅ + …………………… = 2^{n – 1}

Question 6.
Prove that for any real numbers a, d, a . C₀ + (a + d) . C₁ + (a + 2d) . C₂ + ……………… + (a + nd) . C_n = (2a + nd) 2^{n – 1} [May ‘98]
Solution:
Let S = a . C₀ + (a + d) . C₁ + (a + 2d) . C₂ + ……………… + (a + nd) . C_n ………………(1)
By writing the terms in R.H.S of (1) in reverse order has done we get
S = (a + nd)C_n + (a + (n – 1)d)C_n-1 + (a + (n-2)d)C_n-2 + …………. + aC₀
S = (a + nd)C₀ + (a + (n – 1)d)C₁ + (a + (2n – 2)d)C₂ + ………….. + aC_n ……………(2)
Adding (1) and (2) we get

2s = (2a + nd) (C₀ + C₁ + C₂ + …………. + C_n)
2s = (2a + nd)²ⁿ
⇒ S = (2a + nd)^2n-1
∴ aC₀ + (a + d)C₁ + (a + 2d)C₂ + …………. + (a + nd)C_n = (2a + nd) 2^2n-1.

Question 7.
For r = 0, 1, 2, ….., n, prove that C₀ . C_r + C₁ . C_r+1 + C₂ . C_r+2 + ……………… + C_n-r . C_n = \({ }^{2 n} C_{(n+r)}\)
and hence deduce that
i) C₀² + C₁² + C₂² + ………….. + C_n² = \({ }^{2 n} C_n\)
ii) C₀ . C₁ + C₁ . C₂ + C₂ . C₃ + ………… + C_n-1 . C_n = \({ }^{2 n} C_{n+1}\)
[May ‘97, 95, ‘90, ‘94,’93, ‘92; March ‘98, ‘93, AP-Mar. 2018; TS – Mar. 2015]
Solution:
We know that
(1 + x)ⁿ = C₀ + C₁ . x + C₂ . x² + ………………….. + C_r x^r + C_n . xⁿ ………………(1)
(x + 1)ⁿ = C₀xⁿ + C₁x^n-1 + C₂x^n-2 + ………………… + C_rx^n-r + …………………. + C_n ………………(2)
Multiplying (2) and (1), we get
(C₀xⁿ + C₁x^n-1 + C₂x^n-2 + ………………… + C_rx^n-r + …………………. + C_n) (C₀ + C₁ . x + C₂ . x² + ………………….. + C_r x^r + C_n . xⁿ)
= (x + 1)ⁿ (1 + x)ⁿ = (1 + x)²ⁿ
Comparing the coefficient of x^n+r on bothsides we get,
C₀ . C_r + C₁ . C_r+1 + C₂ . C_r+2 + ……………… + C_n-r . C_n = \({ }^{2 n} C_{(n+r)}\) ……………….(3)

i) On substituting r = 0 in equation (3) we get
C₀ . C₀ + C₁ . C₁ + C₂ . C₂ + ……………….. + C_n . C_n = \({ }^{2 n} C_n\)
C₀² + C₁² + C₂² + ………….. + C_n² = \({ }^{2 n} C_n\)

ii) On substituting, r = 1 in equation (3) we get
C₀ . C₁ + C₁ . C₂ + C₂ . C₃ + ………… + C_n-1 . C_n = \({ }^{2 n} C_{n+1}\)

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

Question 9.
If n is a positive integer, then prove that
C₀ + \(\frac{C_1}{2}+\frac{C_2}{3}+\ldots .+\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\) [TS – Mar. 2017]
Solution:
Let S = C₀ + \(\frac{\mathrm{C}_1}{2}+\frac{\mathrm{C}_2}{3}+\ldots . .+\frac{\mathrm{C}_{\mathrm{n}}}{\mathrm{n}+1}\)
= \({ }^{\mathrm{n}} \mathrm{C}_0+\frac{1}{2} \cdot{ }^{\mathrm{n}} \mathrm{C}_1+\frac{1}{3} \cdot{ }^{\mathrm{n}} \mathrm{C}_2+\frac{1}{\mathrm{n}+1} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}\)
Now multiplying on both sides with (n + 1), we get

Question 10.
Prove 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].
Solution:

Question 11.
If (1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ……………… + a_2nx²ⁿ, then prove that
i) a₀ + a₁ + a₂ + …………….. + a_2n = 3ⁿ
ii) a₀ + a₂ + a₄ + ……………. + a_2n = \(\frac{3^n+1}{2}\)
iii) a₁ + a₃ + a₅ + ……………… + a_2n = \(\frac{3^n-1}{2}\)
iv) a₀ + a₃ + a₆ + a₉ + ………………. = 3^n-1 [TS May 2016; Board Paper]
Solution:
(1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ……………… + a_2nx²ⁿ
Put x = 1 in equation (1) we get,
(1 + 1 + 1²)ⁿ = a₀ + a₁ (1) + a₂ (1)² + ……………… + a_2n. (1)²ⁿ
a₀ + a₁ + a₂ + …………… + a_2n = 3ⁿ ……………(2)
Put x = – 1 in equation (1) we get,
(1 – 1 + 1²)ⁿ = a₀ + a₁ (- 1) + a₂ (- )² + a₃. (- 1)⁴ ……………… + a_2n. (- 1)²ⁿ
a₀ – a₁ + a₂ – a₃ + a₄ + …………… + a_2n = 1 ……………(3)
i) From (2),
a₀ + a₁ + a₂ + a₃ + …………… + a_2n = 3ⁿ

ii) Now, adding (2) and (3)
a₀ + a₁ + a₂ + a₃ + …………… + a_2n = 3ⁿ

iii) Now, subtracting (2) and (3)

iv) Put, x = 1 in equation (1) we get
(1 + 1 + 1²)ⁿ = a₀ + a₁ (1) + a₂(1)² + a₃ (1)³ + a₄ (1)⁴ + a₅ (1)⁵ + a₆ (1)⁶ + ……………… + a_2n (1)²ⁿ
= a₀ + a₁ + a₂ + a₃ + a₄ + a₅ + a₆ + …………….. + a_2n = 3ⁿ ………………(4)

Put x = ω in equation (1), we get
(1 + ω + ω²)ⁿ = a₀ + a₁ (ω) + a₂ (ω)² + a₃ (ω)³ + a₄ (ω)⁴ + a₅ (ω)⁵+ a₆ (ω)⁶ + ……………. + a_2n (ω)²ⁿ
= a₀ + a₁ . (ω) + a₂ (ω)² + a₃ (ω)³ + a₄ (ω)⁴ + a₅ (ω)⁵ + a₆ (ω)⁶ + ……………. + a_2n (ω)²ⁿ = 0 …………….(5)
put x = in equation (1) we get,
(1 + ω² + ω⁴)ⁿ = a₀ + a₁ . (ω)² + a₂ (ω)⁴ + a₃ (ω)⁶ + a₄ (ω)⁸ + a₅ (ω)¹⁰ + a₆ (ω)¹² + ……………. + a_2n (ω²)²ⁿ
= a₀ + a₁ . ω² + a₂ ω + a₃ + a₄ ω² + a₅ ω + a₆ + ……………. + a_2n (ω²)²ⁿ = 0
Now, adding (4), (5) and (6)

⇒ 3a₀ + 0 + 0 + 3a₃ + 0 + 0 + 3a₆ + ………….. = 3ⁿ
⇒ 3(a₀ + a₃ + a₆ + ………………..) = 3ⁿ
⇒ a₀ + a₃ + a₆ + ……………….. = 3^n-1.

Question 12.
If (1 + x + x² + x³)⁷ = b₀ + b₁x + b₂x² + ………………. + b₂₁x²¹, then find the value of
i) b₀ + b₂ + b₄ + …………. + b₂₀
ii) b₁ + b₃ + b₅ + …………….. + b₂₁
Solution:
Given
(1 + x + x² + x³)⁷ = b₀ + b₁x + b₂x² + ………………. + b₂₁x²¹ …………………….(1)
Substituting x = 1 in (1),
we get b₀ + b₁ + b₂ + …………. + b₂₁ = 47 ………………(2)
Substituting x = – 1 in (1),
we get b₀ – b₁ + b₂ + ……………… – b₂₁ = 0 ……………..(3)
i) (2) + (3)
⇒ 2b₀ + 2b₂ + 2b₄ + ………………….. + 2b₉₀ = 4⁷
⇒ b₀ + b₂ + b₄ + …………….. + b₂₀ = 2¹³

ii) (2) – (3)
⇒ 2b₁ + 2b₃ + 2b₅ + ……………. + 2b₂₁ = 4⁷
⇒ b₁ + b₃ + b₅ + ……………. + b₂₁ = 2¹³

Question 13.
the coefficients of x⁹, x¹⁰, x¹¹ in the expansion of (1 + x)ⁿ are in A.P., then prove that n² – 41n + 398 = 0.
Solution:
Coefficient of r in the expansion of (1 + x)ⁿ is \({ }^n C_r\).
Given coefficients of x⁹, x¹⁰, x¹¹ in the expansion of (1 + x)ⁿ are \({ }^n C_9,{ }^n C_{10},{ }^n C_{11}\) in AP., then 2\(\left({ }^n C_{10}\right)={ }^n C_9+{ }^n C_{11}\)
⇒ \(2 \frac{n !}{(n-10) ! 10 !}=\frac{n !}{(n-9) ! 9 !}+\frac{n !}{(n-11) !+11 !}\)
⇒ \(\frac{2}{10(n-10)}=\frac{1}{(n-9)(n-10)}+\frac{1}{11 \times 10}\)
⇒ \(\frac{2}{(n-10) 10}=\frac{110+(n-9)(n-10)}{110(n-9)(n-10)}\)
⇒ 22 (n – 9) = 110 + n² – 19n + 90
⇒ n² – 41n + 398 = 0.

Question 14.
1f 36, 84, 126 are three successive binomial coefficlenü* in the expansion of (1 + x)ⁿ, then find n. [May ’09, ’06] [TS – Mar. 2019]
Solution:
We know that
(1 + x)ⁿ = C₀ + C₁x + C₂x² + ………………. C_nxⁿ.
Let the three successive binomial coefficients in the expansion of (1 + x)ⁿ are \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1},{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}},{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\)
Given that,
\({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}\) = 36 ………………(1)
\({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) = 84 ………………(2)
\({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}\) = 126 ……………..(3)

⇒ 2n – 2r = 3r + 3
⇒ 5r = 2n – 3 from (4)
⇒ 5 \(\left(\frac{3 n+3}{10}\right)\) = 2n – 3
⇒ 3n + 3 = 4n – 6
⇒ n = 9.

Question 15.
If the 2nd, 3rd and 4th terms in the expansion of (a + x)ⁿ are respectively 240, 720, 1080, find a, x, n. [TS – Mar. 2016; Ma ‘09, ’06]
Solution:
The general term in the expansion of (x + a)ⁿ is
T_r+1 = x^n-r . a^r
In the expansion, (a + x)ⁿ
T₂ = T₁₊₁
= \({ }^n C_1\) (a)^n-1 x¹ = 240 …………..(1)
T₃ = T₂₊₁
= \({ }^n C_2\) (a)^n-2 x² = 720 …………..(2)
T₄ = T₃₊₁
= \({ }^n C_3\) (a)^n-3 x³ = 1080 …………..(3)

⇒ 2(n – 2) x = 9a ………………(5)
\(\frac{(5)}{(4)}=\frac{2(n-2) x}{(n-1) x}=\frac{9 a}{6 a}\)
⇒ 4(n – 2) = 3(n – 1)
⇒ 4n – 8 = 3n – 3
⇒ n = 5
Substitute n = 5 in equation (4)
⇒ (5 – 1)x = 6a
⇒ 4x = 6a
⇒ 2x = 3a
⇒ x = \(\frac{3 a}{2}\)
Substitute n, x in equation (1).
\(5 C_1(a)^{5-1}\left(\frac{3 a}{2}\right)^1\) = 240
⇒ 5 . a⁴ . \(\frac{3}{2}\) . a = 240
⇒ a⁵ = 32
⇒ a = 2.
x = \(\frac{3 a}{2}\)
x = 3 . \(\frac{2}{2}\)
⇒ x = 3
∴ a = 2, x = 3, n = 5.

Question 16.
If the coefficients of rth, (r + 1)th and (r + 2)nd tenus in the expansion of (1 + x)ⁿ are in A.P then show that n² – (4r + 1) n + 4r² – 2 = 0.
Solution:
In the expansion (1 + x)ⁿ
The coeff. of T_r = \({ }^n \mathrm{C}_{\mathrm{r}-1}\);
The coeff. of T_r+1 = \({ }^n C_r\)
The coeÍf. of T_r+2 = \({ }^n \mathrm{C}_{\mathrm{r}+1}\)
Given that
\({ }^n \mathrm{C}_{\mathrm{r}-1}\), \({ }^n C_r\), \({ }^n \mathrm{C}_{\mathrm{r}+1}\) are in AP.
If a, b, c are in A.P.
Then 2b = a + c

2(n – r + 1) (r + 1) = (r + 1) r + (n – r + 1) (n – r)
⇒ 2nr + 2n – 2r² – 2r + 2r + 2 = r² + r + n² – nr – nr + r² + n – r
⇒ 2nr + 2n – 2r² + 2 = n² – 2nr + n + 2r²
⇒ n² – 2nr + n + 2r² – 2nr – 2n + 2r² – 2 = 0
⇒ n² – 4nr + 4r² – n – 2 = 0
⇒ n² – (4r + 1)n + 4r² – 2 = 0.

Question 17.
If ‘P’ and ‘Q’ are the sum of odd tenns and the sum of even terms respectively in the expansion of (x + a)ⁿ then prove that
i) P² – Q² = (x² – a²)ⁿ
ii) 4PQ = (x + a)²ⁿ – (x – a)²ⁿ [AP – March 2016, March ’10]
Solution:
Since P and Q are the sum of odd terms and sum of even terms respectively in the expansion of (x + a)ⁿ.
Now (x + a)ⁿ =

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

ii) L.H.S : 4PQ = (P + Q)² – (P – Q)²
= ((x + a)ⁿ)² – ((x – a)ⁿ)²
= (x + a)²ⁿ – (x – a)²ⁿ = R.H.S.
∴ 4PQ = (x + a)²ⁿ – (x – a)²ⁿ.

Question 18.
If the coefficients of 4 consecutive term in the expansion of (1 + x)ⁿ are a₁, a₂, a₃, a₄ 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’11. ‘07, March ’11, ’95].
Solution:
Since a₁, a₂, a₃, a₄, are the coefficients of 4 consecutive terms in the expansion of (1 + x)ⁿ
Let a₁ = \({ }^n C_{r-1}\),
a₂ = \({ }^n C_r\),
a₃ = \({ }^n C_{r+1}\),
a₄ = \({ }^n C_{r+2}\)
L.H.S:

R.H.S:

Question 19.
Show that the middle term in the expansion of (1 + x)²ⁿ is \(\frac{1 \cdot 3 \cdot 5 \ldots \ldots(2 n-1)}{n !}(2 x)^n\).
Solution:
The expansion of (1 + x)²ⁿ contains 2n + 1 terms.
∴ Middle term is (n + 1)th term
∴ T_n+1 = \({ }^{2 n} C_n\) xⁿ

= \(\frac{2 n !}{n ! \cdot n !}\) xⁿ

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

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

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

∴ Middle term = \(\frac{1 \cdot 3 \cdot 5 \ldots \ldots(2 n-1)}{n !}(2 x)^n\).

Question 20.
If (1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + ……………. + ax²⁰ then prove that
i) a₀ + a₁ + a₂ + ………….. + a₂₀ = 2¹⁰
ii) a₀ – a₁ + a₂ – a₃ + ……………. + a₂₀ = 4¹⁰
Solution:
Given
(1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + ……………. + ax²⁰ …………….(1)

i) Put x = 1, in (1), we get
(1 + 3 – 2)¹⁰ = a₀ + a₁ + a₂ + ………….. + a₂₀
⇒ a₀ + a₁ + a₂ + …………….. + a₂₀ = 2¹⁰.

ii) Put x = – 1 in (1), we get
(1 – 3 – 2)’°= a₀ – a₁ + a₂ – a₃ + ………………. + a₂₀
⇒ a₀ – a₁ + a₂ – a₃ + ……………. + a₂₀ = 4¹⁰.

Question 21.
If 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]
Solution:
Now, \(\frac{{ }^{n^n} C_r}{{ }^{n_C} C_{r-1}}=\frac{\frac{n !}{(n-r) ! r !}}{\frac{n !}{(n-r+1) !(r-1) !}}\)
= \(\frac{(n-r+1) !(r-1) !}{(n-r) ! r !}\)

Question 22.
Find 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]
Solution:
Let the given series ¡s
S = 1 + \(\frac{2}{3} \cdot \frac{1}{2}+\frac{2 \cdot 5}{3 \cdot 6} \cdot \frac{1}{2^2}+\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9} \cdot \frac{1}{2^3}+\cdots \cdots\)
Comparing with 1 + nx + \(\frac{n(n-1)}{1 \cdot 2}\) x² + …………….

Question 23.
\(\frac{3 \cdot 5}{5 \cdot 10}+\frac{3 \cdot 5 \cdot 7}{5 \cdot 10 \cdot 15}+\frac{3 \cdot 5 \cdot 7 \cdot 9}{5 \cdot 10 \cdot 15 \cdot 20}+\cdots \cdots \cdots \infty\) [TS – Mar. 2017; May ’09]
Solution:
Let the given series is
S = \(\frac{3 \cdot 5}{5 \cdot 10}+\frac{3 \cdot 5 \cdot 7}{5 \cdot 10 \cdot 15}+\frac{3 \cdot 5 \cdot 7 \cdot 9}{5 \cdot 10 \cdot 15 \cdot 20}+\cdots \cdots \cdots \infty\)

S + \(\frac{3}{5}\) = \(\frac{3}{5}+\frac{3 \cdot 5}{5 \cdot 10}+\frac{3 \cdot 5 \cdot 7}{5 \cdot 10 \cdot 15}+\frac{3 \cdot 5 \cdot 7 \cdot 9}{5 \cdot 10 \cdot 15 \cdot 20}+\ldots \ldots\)

Question 24.
If 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 + 6x. [TS – Mar. 2016, May ’14, ’11, ’08, Mar. ’14, ’12, ’06, TS – Mar. ’19]
Solution:
Given x = \(\frac{1}{5}+\frac{1 \cdot 3}{5 \cdot 10}+\frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15}+\ldots \ldots \ldots\),
x + 1 = 1 + \(\frac{1}{5}+\frac{1 \cdot 3}{5 \cdot 10}+\frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15}+\cdots \cdots\)
Now, comparing with

⇒ x² + 1 + 2x = \(\frac{5}{3}\)
⇒ 3x² + 6x + 3 = 5
⇒ 3x² + 6x = 2.

Question 25.
Find 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]
Solution:
Let the given series is
S = \(\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots \ldots \ldots\)
S + 1 = 1 + \(\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots \ldots \ldots\)
Now, comparing with

Question 26.
If 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 + 24x = 11. [AP – MAr. ’19, ’17, ’15, May ’16; TS – MAr. ’18, ’16, Mar. ’09, Board Paper, May ’15]
Solution:
Given

Question 27.
If 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 + 4x. [TS – May; March ’13, May ’12]
Solution:
Given,

Question 28.
Find 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].
Solution:

Question 29.
Show that for any non-zero rational number x.
1 + \(\frac{x}{2}+\frac{x(x-1)}{2 \cdot 4}+\frac{x(x-1)(x-2)}{2 \cdot 4 \cdot 6}+\ldots \ldots\) = 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]
Solution:
L.H.S = 1 + \(\frac{\mathrm{n}}{2}+\frac{\mathrm{n}(\mathrm{n}-1)}{2 \cdot 4}+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{2 \cdot 4 \cdot 6}+\ldots\)

Some More Maths 2A Binomial Theorem Important Questions

Question 1.
Find the 7th term in the expansion of \(\left(\frac{4}{x^3}+\frac{x^2}{2}\right)^{14}\).
Solution:
Given \(\left(\frac{4}{x^3}+\frac{x^2}{2}\right)^{14}\)
Here, x = \(\frac{4}{x^3}\); a = \(\frac{x^2}{2}\); n = 14
r + 1 = 7
⇒ r = 6
The general term in the expansion of (x + a)ⁿ is
T_r+1 = \({ }^n C_r\) x^n-r a^r
The 7th term in the expansion of \(\left(\frac{4}{x^3}+\frac{x^2}{2}\right)^{14}\)

Question 2.
Find the 5th term in the expansion of (3x – 4y)⁷.
Solution:
Given (3x – 4y)⁷
Here, x = 3x; a = – 4y; n = 7
r + 1 = 5
r =4
The general term in the expansion of (x + a)ⁿ is
T_{r + 1} = \({ }^n C_r\) x^n-r a^r
The rth term in the expansion of (3x – 4y)⁷ is
T₄₊₁ = \({ }^7 \mathrm{C}_4\) (3x)^7-4 (- 4y)⁴
T₅ = \({ }^7 \mathrm{C}_4\) (3x)³ (4y)⁴
= 35 . 3³ . x³ 4⁴ . y⁴
= 241920 x³ . y⁴

Question 3.
Find the middle term(s) in \(\left(\frac{3}{a^3}+5 a^4\right)^{20}\).
Solution:
Given \(\left(\frac{3}{a^3}+5 a^4\right)^{20}\)
Here, x = \(\frac{3}{a^3}\), a = 5a⁴, n = 20
Since, n = 20 is even,
Middle term = \(\frac{n}{2}\) + 1
= \(\frac{20}{2}\) + 1
= 10 + 1 = 11th term
r + 1 = 11
⇒ r = 10
The general term in this expansion is
T_r+1 = \({ }^{{ }^n} C_r\) x^n-r a^r
11th term of given expansion is
T₁₀₊₁ = \({ }^{20} C_{10}\left(\frac{3}{a^3}\right)^{20-10}\left(5 a^4\right)^{10}\)
= \({ }^{20} \mathrm{c}_{10}\left(\frac{3}{a^3}\right)^{10}\left(5 a^4\right)^{10}\)

T₁₁ = \({ }^{20} C_{10}\) . a^-30 . 5¹⁰ . a⁴⁰
= \({ }^{20} C_{10}\) . 3¹⁰ . 5¹⁰ . a¹⁰

Question 4.
Find the middle term(s) in the expansion of (4x² + 5x³)¹⁷.
Solution:
Given (4x² + 5x³)¹⁷
Here, x = 4x², a = 5x³, n = 17
Since, n = 17 is odd, then
Middle terms \(\frac{\mathrm{n}+1}{2}, \frac{\mathrm{n}+3}{2}\) = 9, 10 terms
9’ term:
r + 1 = 9
⇒ r = 8
The general term in this expansion is
T_r+1 = \({ }^{{ }^n} C_r\) x^n-r a^r
9th term of given expansion is
T₈₊₁ = \({ }^{17} \mathrm{c}_9\) (4x²)^17-8 (5x³)⁸
= \({ }^{17} \mathrm{c}_9\) (4x²)⁹ (5x³)⁸
= \({ }^{17} \mathrm{c}_9\) . 4⁹ . x¹⁸ . 5⁸ . x²⁴
= \({ }^{17} \mathrm{c}_9\) . 4⁹ . 5⁸ . x⁴²

10th term:
r + 1 = 10
⇒ r = 9
The general term in this expansion is
T_r+1 = \({ }^{{ }^n} C_r\) x^n-r a^r
10th term in this expansion ¡s
T₉₊₁ = \({ }^{17} \mathrm{C}_9\)(4x²)^17-9 (5x³)⁹
T₁₀ = \({ }^{17} \mathrm{C}_9\) (4x²)⁸ (5x³)⁹
= \({ }^{17} \mathrm{C}_9\) . 4⁸ . 5⁹ . x¹⁶ . x²⁷
= \({ }^{17} \mathrm{C}_9\) . 4⁸ . 5⁹ . x⁴³

Question 5.
Find the coefficient of x¹¹ in \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\).
Solution:
Given \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\)
Here, x = 2x²,
a = \(\frac{3}{x^3}\), n = 13
Now, the general term in this expansion is
T_r+1 = \(\) x^n-r a^r
= \({ }^{13} \mathrm{C}_{\mathrm{r}}\) (2x^r)^13-r \(\left(\frac{3}{x^3}\right)^r\)
= \({ }^{13} \mathrm{C}_{\mathrm{r}}\) 2^13-r x^26-2r . 3² . x^-3r
= \({ }^{13} \mathrm{C}_{\mathrm{r}}\) 2^13-r 3^r x^26-5r ………………(1)
To find the coefficient of x¹¹
Put 26 – 5r = 11
5r = 15
⇒ r = 3
Now substituting r = 3 in equation (1) we get
T₃₊₁ = \({ }^{13} \mathrm{C}_3\) 2^13-3 3³ x^26-15
T₄ = \({ }^{13} \mathrm{C}_3\) 2¹⁰ 3³ x¹¹
The coefficient of x¹¹ in the expansion of \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\) is \({ }^{13} \mathrm{C}_3\) 2¹⁰ 3³.

Question 6.
Find the term independent of x in the expansion of \(\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^2}\right)^{10}\).
Solution:
Given \(\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^2}\right)^{10}\)
Here, x = \(\sqrt{\frac{x}{3}}\), a = \(\frac{3}{2 x^2}\), n = 10
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
T_r+1 = \({ }^{10} C_r\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\frac{3}{2 x^2}\right)^r\)
= \({ }^{10} C_r \frac{x^{\frac{10-r}{2}}}{3^{\frac{10-r}{2}}} \cdot \frac{3^r}{2^r} \cdot x^{-2 r}\)
= \({ }^{10} \mathrm{C}_{\mathrm{r}} \frac{3^{\mathrm{r}}}{3^{\frac{10-\mathrm{r}}{2}} \cdot 2^{\mathrm{r}}} \mathrm{x}^{\frac{10-\mathrm{r}}{2}-2 \mathrm{r}}\) ……………….(1)
To find the term independent of x¹¹ (i.e., coeff. of x°)
Put \(\frac{10-\mathrm{r}}{2}\) – 2r = 0
10 – r – 4r = 0
5r = 10
⇒ r = 2
Substitute r = 2 in equation (1) we get
T = \({ }^{10} \mathrm{C}_2 \frac{3^2}{3^{\frac{10-2}{2}} \cdot 2^2} \cdot \mathrm{x}^{\frac{10-2}{2}-2^2}\)
= \({ }^{10} C_2 \frac{3^2}{3^3 \cdot 2^2} \cdot x^0\)
= \(\frac{10 \cdot 9}{1 \cdot 2} \times \frac{3^2}{3^4 \cdot 2^2}=\frac{10 \cdot 9}{1 \cdot 2} \cdot \frac{1}{9 \cdot 4} \cdot x^0=\frac{5}{4} x^0\)
The term independent of x in the given expansion is T₃ = \(\frac{5}{4}\).

Question 7.
Find the largest binomial coefficients In the expansion of (1 + x)¹⁹.
Solution:
Given, (1 + x)¹⁹
Here n = 19, an odd integer.
The largest binomial coefficients are \({ }^n C_{\frac{n-1}{2}}\) and \({ }^n C_{\frac{n+1}{2}}\)
\({ }^{19} \mathrm{C}_{\frac{19-1}{2}} \text { and }{ }^{19} \mathrm{C}_{\frac{19+1}{2}}\) = \({ }^{19} C_9 \text { and }{ }^{19} C_{10}\)
(Note that, \({ }^{19} C_{9}={ }^{19} C_{10}\)).

Question 8.
If the coefficients of (2r + 4)th, (r – 2)th terms in the expansion of (1 + x)²¹ equal find ‘r’.
Solution:
Given (1 – x)¹⁸
The general term in the expansion of (x + a)ⁿ is
T_r+1 = \({ }^n C_r\) x^n-r a^r
(2r + 4)th term in the expansion of (1 + x)¹⁸ is
T_(2r+3)+1 = \({ }^{18} \mathrm{C}_{2 \mathrm{r}+3}\) (1) x^2r+3
T_2r+4 = \({ }^{18} \mathrm{C}_{2 \mathrm{r}+3}\) x^2r+3
The coeff. of (2r + 4)th term is \({ }^{18} \mathrm{C}_{2 \mathrm{r}+3}\)
(r – 2) term in the expansion of (1 + x)¹⁸ is
T_(r-3)+1 = \({ }^{18} \mathrm{C}_{\mathrm{r}-3}\) 1^18-(r-3) x^r-3
T_r-2 = \({ }^{18} \mathrm{C}_{\mathrm{r}-3}\) . x^r-3
The coeff. of (r – 2)nd term is \({ }^{18} \mathrm{C}_{\mathrm{r}-3}\)
∴ Given that two coefficients are equal.
\({ }^{18} \mathrm{C}_{2 \mathrm{r}+3}={ }^{18} \mathrm{C}_{\mathrm{r}-3}\)
\({ }^n C_r={ }^n C_s\)
⇒ n = r + s (or) r = s
18 = 2r + 3 + r – 3
2r + 3 = r – 3
3r = 18
⇒ r = 6
⇒ r = – 6
Since, r is a positive integer, we get r = 6.

Question 9.
Find the set E of x for which the binomial expansion (7 + 3x)^-5.
Solution:
Given (7 + 3x)^-5
7^-5 (1 + \(\frac{3 x}{7}\))^-5
The binomial expansion of (7 + 3x)^-5 is valid when
\(\left|\frac{3 x}{7}\right|<1 \Rightarrow|x|<\frac{7}{3}\)
x ∈ \(\left(\frac{-7}{3}, \frac{7}{3}\right)\)

Question 10.
Find the set E of x for which the binomial expansion (7 – 4x)^-5 is valid.
Solution:
Given (7 – 4x)^-5 = 7^-5 (1 – \(\frac{4 x}{7}\))^-5
The binomial expansion of (7 – 4x)^-5 is valid when
\(\left|\frac{-4 x}{7}\right|<1 \Rightarrow\left|\frac{4 x}{7}\right|<1\)
⇒ x < \(\frac{7}{4}\)
⇒ x ∈ \(\left(\frac{-7}{4}, \frac{7}{4}\right)\)
∴ E = \(\left(\frac{-7}{4}, \frac{7}{4}\right)\)

Question 11.
Find the 6th term of (1 + \(\frac{x}{2}\))^-5
Solution:
Given, (1 + \(\frac{x}{2}\)))^-5
Comparing this with (1 + x)ⁿ, where
x = \(\frac{x}{2}\), n = – 5,
⇒ r + 1 = 6
⇒ r = 5
The general term in the binomial expansion of (1 + x)ⁿ is

Question 12.
Find the coefficient of x⁶ in (3x – \(\frac{4}{x}\))¹⁰.
Solution:
Given (3x – \(\frac{4}{x}\))¹⁰
Here, x = 3x; a = – \(\frac{4}{x}\), n = 10
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
= \({ }^{10} \mathrm{C}_{\mathrm{r}}\) (3x)^10-r \(\left(\frac{-4}{x}\right)^{\mathbf{r}}\)
= \({ }^{10} \mathrm{C}_{\mathrm{r}}\) 3^10-r x^10-r (- 4)^r x^r
= \({ }^{10} \mathrm{C}_{\mathrm{r}}\) 3^10-r . (- 4)^10-r . x^10-2r …………..(1)
To find the coefficient of x^-6,
put 10 – 2r = – 6
⇒ 2r = 16
⇒ r = 8
Now, substituting r = 8 in equation (1) we get
T₈₊₁ = \({ }^{10} \mathrm{C}_8\) 3^10-8 (- 4)⁸ x^10-16
T₉ = \({ }^{10} \mathrm{C}_8\) 3² (4)⁸ x^-6
The coeff. of x^-6 in the expansion of (3x – \(\frac{4}{x}\))¹⁰ is \({ }^{10} \mathrm{C}_8\) . 3² . (4)⁸

Question 13.
Find the middle term (s) in the expansion of \(\left(4 a+\frac{3}{2} b\right)^{11}\).
Solution:
Given \(\left(4 a+\frac{3}{2} b\right)^{11}\)
Here, x = 4a; a = \(\frac{3}{2}\) b; n = 11
Since, n = 11 is odd,
Middle terms = \(\frac{\mathrm{n}+1}{2}, \frac{\mathrm{n}+3}{2}\) = 6, 7 terms.

6th term :
r + 1 = 6
⇒ r = 5
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
6th term of given expansion ¡s
T₅₊₁ = \({ }^{11} C_5(4 a)^{11-5}\left(\frac{3}{2} b\right)^5\)
T₆ = \({ }^{11} \mathrm{C}_5(4 \mathrm{a})^6\left(\frac{3}{2} \mathrm{~b}\right)^5\)
= \(=11 C_5 \frac{4^6 \cdot 3^5}{2^5} \cdot a^6 \cdot b^5\)
T₆ = \({ }^{11} \mathrm{C}_6 \frac{2^{12} \cdot 3^5}{2^5}\) . a⁴b⁵
= \({ }^{11} \mathrm{C}_6\) . a⁶ . b⁵

7th term :
r + 1 = 7
⇒ r = 6
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
7th term of given expansion is

Question 14.
Find the middle term (s) in the expansion of (4x² + 5x³)¹⁷.
Solution:
Given (4x² + 5x³)¹⁷.
Here, x = 4x², a = 5x³, n= 17.
Since, n = 17 is odd, then
middle terms = \(\frac{n+1}{2}, \frac{n+3}{2}\) = 9, 10 terms

9th term :
r + 1 = 9
⇒ r = 8
The general term in this expansion is
T_r+1 = \({ }^n C_r\) x^n-r a^r
∴ 9th term in this expansion is
T₉₊₁ = \({ }^{17} \mathrm{C}_9\)(4x²)^17-9 (5x³)⁹
T₁₀ = \({ }^{17} \mathrm{C}_9\) (4x²)⁸ (5x³)⁹
= \({ }^{17} \mathrm{C}_9\) 4⁸ . 5⁹ x¹⁶ . x²⁷
= \({ }^{17} \mathrm{C}_9\) 4⁸ . 5⁹ . x⁴³

Question 15.
Prove that 2 . C₀ + 5 . C₁ + 8 . C₂ + ……………… + (3n + 2) C_n = (3n + 4) 2^n-1
Solution:
Let S = 2 . C₀ + 5 . C₁ + 8 . C₂ + ……………… + (3n + 2) C_n = (3n + 4) 2^n-1 …………….(1)
By writing the terms in (1) in the reverse order
S = (3n + 2) . C_n + (3n – 1) . C_n-1 + (3n – 4) C_n-2 + ……………….. + 2. C₀
S = (3n + 2) C₀ + (3n – 1) C₁ + (3n – 4) C₂ + ……………… + 2 C_n ………….. (2)
Adding (1) and (2) we get,
S = 2 . C₀ + 5 . C₁ + 8 . C₂ + …………… + (3n + 2) . C_n
S = (3n + 2) . C₀ + (3n – 1) . C₁ + (3n – 4) . C₂ + ……………. + 2 . C_n
2S = (3n + 4)C₀ + (3n + 4) C₁ . (3n + 4) C₂ + …………… + (3n + 4) C_n
2S = (3n + 4) (C₀ + C₁ + C₂ + …………….. + C_n)
2S = (3n + 4) 2ⁿ
⇒ S = (3n + 4) 2^n-1
∴ 2 . C₀ + 5 . C₁ + 8. C₂ + …………… + (3n + 2) C_n = (3n + 4) . 2^n-1

Question 16.
Prove that
(C₀ + C₁) (C₁ + C₁) (C₂ + C₃) + ………………… + (C_n-1 + C_n) = \(\frac{(n+1)^n}{n !}\) . C₀ . C₁ . C₂ ……………. C_n

Question 17.
Write the first 3 terms in the expansion of \(\left(1+\frac{\mathrm{x}}{2}\right)^{-5}\).
Solution:
Given \(\left(1+\frac{\mathrm{x}}{2}\right)^{-5}\)
We know that,

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

Question 18.
Find the coefficient of x⁶ in the expansion of (1 – 3x)^-2/5.
Solution:
Given that (1 + 3x)^-2/5
The general term of (1 – 3x)^-2/5 is
T_r+1 = \(\frac{n(n-1)(n-2) \ldots .(n-r+1)}{r !}\)
where x = – 3x, n = \(\frac{-2}{5}\)

T_r+1 = \(\frac{\frac{-2}{5}\left(\frac{-2}{5}-1\right)\left(\frac{-2}{5}-2\right) \ldots \ldots . .\left(\frac{-2}{5}-r+1\right)}{r !}(-3 x)^r\)

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

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

Put r = 6
Then the coefficient of x^-6 is \(\frac{2 \cdot 7 \cdot 12 \ldots \ldots .27}{5^6 \cdot 6 !}(3)^6\)
= \(\frac{2 \cdot 7 \cdot 12 \ldots \ldots .27}{6 !} \cdot\left(\frac{3}{5}\right)^6\).

Question 19.
Eind the coefficient of x⁴ in the expansion of (1 – 4x)^3/5
Solution:
Given (1 – 4x)^3/5
The general term of (1 – 4x)^3/5 is
T_r+1 = \(\frac{n(n-1)(n-2) \ldots \ldots(n-r+1)}{r !}\) . x^r
where, x = – 4x, n = \(\frac{-3}{5}\)

Question 20.
Find 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]
Solution:
Let the given series is
S = 1 + \(\frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\ldots \ldots\)
Now, comparing with

Question 21.
Find the sum of the infinite series 1 – \(\frac{4}{5}+\frac{4 \cdot 7}{5 \cdot 10}-\frac{4 \cdot 7 \cdot 10}{5 \cdot 10 \cdot 15}\) + …………..
Solution:
Let the given series is
S = 1 – \(\frac{4}{5}+\frac{4 \cdot 7}{5 \cdot 10}-\frac{4 \cdot 7 \cdot 10}{5 \cdot 10 \cdot 15}\) + …………..
Now comparing with

Question 22.
Find the sum of the infinite series \(\frac{3}{4 \cdot 8}-\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}\).
Solution:
Let the given series

Question 23.
If t = \(\frac{4}{5}+\frac{4 \cdot 6}{5 \cdot 10}+\frac{4 \cdot 6 \cdot 8}{5 \cdot 10 \cdot 15}\) + ……………….. ∞, then prove that 9t = 16.
Solution:
Given

Here p = 4; p + q = 6;
⇒ q = 2
⇒ x = \(\frac{2}{5}\)
∴ (1) ⇒ 1 + t = \(\left(1-\frac{2}{5}\right)^{\frac{-4}{2}}\)
⇒ 1 +t = \(\left(\frac{3}{5}\right)^{-2}\)
⇒ 1 + t = \(\frac{25}{9}\)
⇒ 9t =16.

Question 24.
If I, n are positive integers, 0 < f < 1 and if (7 + 4√3)ⁿ = 1 + f, then show that
i) I is an odd integer and
ii) (I + f) (1 – f) = 1.
Solution:
Sol. Since I, n are positive integers 0 < F < 1 and (7 + 4√3)ⁿ = 1 + F
Now, 36 < 48 < 49
⇒ 6 < 4√3 < 7 ⇒ – 6 > – 4√3 > – 7
⇒ 1 > 7 – 4√3 > 0
⇒ 0 < 7 – 4√3 < 1
⇒ 0< 7 – 4√3 < 1
⇒ o < (7 – 4√3)ⁿ < 1
Let (7 – 4√3)ⁿ = f
∴ 0 < f < 1
Now, 1 + f = (7 + 4√3)ⁿ

where k is an integer
⇒ I + F + f = Even integer.
Since, I is an integer, F + f is an integer.
But, 0 < F < 1, 0 < f < 1
⇒ 0 < F + f < 2.
∴ f + F = 1
From (1)
⇒ I + 1 = Even integer
⇒ Even integer – 1 = Odd Integer
∴ I is an odd integer.

ii) L.H.S: (I + f) (I – f) = (7 + 4√3)ⁿ . F
= (7 + 4√3)ⁿ (7 – 4√3)ⁿ
= (49 – 48)ⁿ = 1ⁿ = 1 = R.H.S
∴ LH.S = R.H.S.

Question 25.
Find the 8th term of \(\left(1-\frac{5 x}{2}\right)^{-3 / 5}\). [AP – Mar. ’18]
Solution:
Given \(\left(1-\frac{5 x}{2}\right)^{-3 / 5}\)
T₈ = \(\frac{\left(\frac{-3}{5}\right)\left(\frac{-3}{5}-1\right)\left(\frac{-3}{5}-2\right) \ldots\left(\frac{-3}{5}-6\right)}{7 !}\left(\frac{-5 x}{2}\right)^7\)
= \(\frac{\left(\frac{3}{5}\right)\left(\frac{8}{5}\right)\left(\frac{13}{5}\right) \ldots\left(\frac{33}{5}\right)}{7 !}\left(\frac{5 x}{2}\right)^7\)
= \(\frac{3 \times 8 \times 13 \times \ldots . \times 33}{7 !}\left(\frac{x}{2}\right)^7\)