{"id":39093,"date":"2022-12-06T10:21:09","date_gmt":"2022-12-06T04:51:09","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=39093"},"modified":"2023-01-18T16:03:33","modified_gmt":"2023-01-18T10:33:33","slug":"maths-1a-mathematical-induction-important-questions","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/maths-1a-mathematical-induction-important-questions\/","title":{"rendered":"TS Inter 1st Year Maths 1A Mathematical Induction Important Questions"},"content":{"rendered":"

Students must practice these Maths 1A Important Questions<\/a> TS Inter 1st Year Maths 1A Mathematical Induction Important Questions to help strengthen their preparations for exams.<\/p>\n

TS Inter 1st Year Maths 1A Mathematical Induction Important Questions<\/h2>\n

Question 1.
\nBy using mathematical induction show that \u2200 n \u2208 N, 12<\/sup> + 22<\/sup> + 32<\/sup> + ……………… + n2<\/sup> = \\(\\frac{n(n+1)(2 n+1)}{6}\\)
\nAnswer:
\nLet S(n) be the statement that 12<\/sup> + 22<\/sup> + 32<\/sup> + …………… + n2<\/sup> = \\(\\frac{n(n+1)(2 n+1)}{6}\\)
\nIf n = 1, then
\nLHS = n2<\/sup> = 12<\/sup> = 1
\n\"TS
\n\u2234 S(k + 1) is true.
\nBy the principle of mathematical induction S(n) is true, \u2200 n \u2208 N.
\n\u2234 12<\/sup> + 22<\/sup> + 32<\/sup> + ……………… +n2<\/sup> = \\(\\frac{n(n+1)(2 n+1)}{6}\\), \u2200 n \u2208 N.<\/p>\n

\"TS<\/p>\n

Question 2.
\nBy using mathematical Induction show that \u2200 n \u2208 N, 13<\/sup> + 23<\/sup> + 33<\/sup> + …………. + n3<\/sup> = \\(\\frac{n^2(n+1)^2}{4}\\). [May 97, 94, 93, 88, Mar. 87]
\nAnswer:
\nLet S(n) be the statement that 13<\/sup> + 23<\/sup> + 33<\/sup> + ……….. + n3<\/sup> = \\(\\frac{n^2(n+1)^2}{4}\\)
\nIf n = 1, then
\nL.H.S = n3<\/sup> = 13<\/sup>
\nR.H.S = \\(\\frac{n^2(n+1)^2}{4}\\) = \\(\\frac{1^2(1+1)^2}{4}\\) = \\(\\frac{1.4}{4}\\) = 1
\n\u2234 L.H.S = R.H.S
\n\u2234 S(1) is true.
\nAssume that S(k) is true.
\n\"TS
\n\u2234 S(k + 1) is true.
\nBy the principle of mathematical induction S(n) is true, \u2200 n \u2208 N.
\n\u2234 13<\/sup> + 23<\/sup> + 33<\/sup> + ………… + n3<\/sup> = \\(\\frac{n^2(n+1)^2}{4}\\), \u2200 n \u2208 N<\/p>\n

Question 3.
\nBy using mathematical induction show that \u2200 n \u2208 N, 2.3 + 3.4 + 4.5 + …………. upto n terms = \\(\\frac{n\\left(n^2+6 n+11\\right)}{3}\\)
\nAnswer:
\n2, 3, 4 ……………………. are in A.P.
\nHere a = 2, d = 3 – 2 = 1
\n\u2234 tn<\/sub> = a + (n – 1)d = 2 + (n – 1)1 = 2 + n – 1 = n + 1.
\n3, 4, 5 …………………… are in A.P.
\nHere a = 3, d = 4 – 3 = 1
\n\u2234 tn<\/sub> = a + (n – 1) d = 3 + (n – 1) 1 = 3 + n – 1 = n + 2.
\n\u2234 The nth<\/sup> term of the given series is (n + 1) (n + 2).
\nLet S(n) be the statement that
\n2.3 + 3.4 + 4.5 + + (n + 1) (n + 2) = \\(\\frac{n\\left(n^2+6 n+11\\right)}{3}\\)
\nIf n = 1, then
\nLHS = (n + 1) (n + 2) = (1 + 1) (1 + 2) = 2.3 = 6
\nRHS = \\(\\frac{\\mathrm{n}\\left(\\mathrm{n}^2+6 \\mathrm{n}+11\\right)}{3}\\) = \\(\\frac{1\\left(1^2+6(1)+11\\right)}{3}\\) = \\(\\frac{18}{3}\\) = 6
\n\u2234 LHS = RHS
\n\u2234 S(1) is true.
\nAssume that S(k) is true.
\n\"TS<\/p>\n

Verification Method:
\n\\(\\frac{\\mathrm{n}\\left(\\mathrm{n}^2+6 n+11\\right)}{3}\\)
\nPut n = k + 1
\n\"TS
\n\u2234 S(k + 1) is true.
\nBy the principle of mathematical induction, S(n) is true \u2200 n \u2208 N.
\n\u2234 2.3 + 3.4 + 4.5 + ……………… + (n + 1) (n + 2) = \\(\\frac{\\mathrm{n}\\left(\\mathrm{n}^2+6 \\mathrm{n}+11\\right)}{3}\\), \u2200 n \u2208 N<\/p>\n

\"TS<\/p>\n

Question 4.
\nBy using mathematical induction show that \u2200 n \u2208 N,
\n\\(\\frac{1}{1.3}+\\frac{1}{3.5}+\\frac{1}{5.7}+\\ldots+\\frac{1}{(2 n-1)(2 n+1)}=\\frac{n}{2 n+1}\\) [Mar. 18 May 15 (AP): May 14, 97, 92]
\nAnswer:
\n\"TS<\/p>\n

Question 5.
\nBy using mathematical induction show that \u2200 n \u2208 N, \\(\\frac{1}{1 \\cdot 4}+\\frac{1}{4 \\cdot 7}+\\frac{1}{7 \\cdot 10}+\\) …………….. upto n terms = \\(\\frac{n}{3 n+1}\\).
\nAnswer:
\n1, 4, 7, ………………… are in A.P.
\nHere, a = 1, d = 4 – 1 = 3
\ntn<\/sub> = a + (n – 1) d = 1 + (n – 1)3 = 1 + 3n – 3 = 3n – 2
\n4, 7, 10, ………………. are in A.P.
\nHere, a = 4, d = 7 – 4 = 3
\ntn<\/sub> = a + (n – 1) d = 4 + (n – 1)3 = 4 + 3n – 3 = 3n + 1
\n\u2234 The nth<\/sup> term in the given series is \\(\\frac{1}{(3 n-2)(3 n+1)}\\).
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 6.
\nBy using mathematical induction show that \u2200 n \u2208 N,
\na + (a + d) + (a + 2d) + upto n terms = \\(\\frac{n}{2}\\) [2a + (n – 1) d].
\nAnswer:
\na, a + d, a + 2d, ………………….. are in A.P
\n\u2234 tn<\/sub> = a + (n – 1) d
\n\u2234 n th term in the given series is a + (n – 1) d.
\nLet S(n) be the statement that
\na + (a + d) + (a + 2d) + ………………. + a + (n – 1) d = \\(\\frac{\\mathrm{n}}{2}\\) [2a + (n – 1) d]
\nIf n = 1 then
\nL.H.S = a + (n – 1)d = a + (1 – 1)d = a
\nR.H.S = \\(\\frac{\\mathrm{n}}{2}\\) [2a + (n – 1)d] = \\(\\frac{1}{2}\\) [2a + (1 – 1)d] = \\(\\frac{1}{2}\\) [2a] = a
\n\u2234 L.H.S = R.H.S
\n\u2234 S(1) is true.
\n[a + (n – 1)d
\nput n = k + 1
\na + (k + 1 – 1)d
\n= a + kd]
\nAssume that S(k) is true.
\na + (a + d) + (a + 2d) + ………………… + [a + (k- 1) d] = \\(\\frac{\\mathrm{k}}{2}\\) [2a + (k – 1)d]
\nAdding (a + kd) on both sides, we get
\n\"TS
\n\u2234 S(k + 1) is true.
\nBy the principle of Mathematical Induction, s(n) is true, \u2200 n \u2208 N
\n\u2234 a + (a + d) + (a + 2d) + ….+ a + (n- 1) d = \\(\\frac{n}{2}\\) [2a + (n – 1) d], \u2200 n \u2208 N<\/p>\n

Question 7.
\nBy using mathematical induction show that \u2200 n \u2208 N,
\na + ar + ar2<\/sup> + ………. upto n terms = \\(\\frac{a\\left(r^n-1\\right)}{r-1}\\), r \u2260 1. [Mar. 19 (AP); Mar. 11, 80; May 87]
\nAnswer:
\na + ar + ar2<\/sup> + ………. are in G.P.
\n\u2234 tn<\/sub> = a . rn – 1<\/sup>
\nThe n th term in the given series is a . r n – 1<\/sup>
\nLet S(n) be the statement that
\na + ar + ar2<\/sup> + …………. + a \u2219 rn – 1<\/sup> = \\(\\frac{a\\left(r^n-1\\right)}{r-1}\\)
\nIf n = 1, then
\nL.H.S = a \u2219 rn – 1<\/sup> = a \u2219 r1 – 1<\/sup> = a \u2219 r0<\/sup> = a \u2219 1 = a
\nR.H.S = \\(\\frac{a\\left(r^n-1\\right)}{r-1}=\\frac{a\\left(r^1-1\\right)}{r-1}=\\frac{a(r-1)}{r-1}\\) = a
\n\u2234 L.H.S = R.H.S
\n\u2234 S(1) is true.
\n[a.rn – 1<\/sup>
\nPut n = k + 1
\na.rk + 1 – 1<\/sup>
\na.rk<\/sup>]
\nAssume that S(k) is true.
\na + ar + ar2<\/sup> + …….. + a.rk – 1<\/sup> = \\(\\frac{a\\left(r^k-1\\right)}{r-1}\\)
\nAdding ark<\/sup> on both sides we get,
\n\"TS
\n\u2234 S(k + 1) is true.
\nBy the principle of mathematical induction, S(n) is true \u2200 n \u2208 N.
\n\u2234 a + ar + ar2<\/sup> + ………………. + a . rn – 1<\/sup> = \\(\\frac{a\\left(r^n-1\\right)}{r-1}\\)<\/p>\n

\"TS<\/p>\n

Question 8.
\nBy using mathematical induction show that \u2200 n \u2208 N, 1.2.3 + 2.3.4 + 3.4.5 + …………………. upto n terms = \\(\\frac{n(n+1)(n+2)(n+3)}{4}\\)
\nAnswer:
\n1, 2, 3 …………………. are in A.P.
\nHere a = 1, d = t2<\/sub> – t1<\/sub> = 2 – 1 = 1
\ntn<\/sub> = a + (n – 1)d = 1 + (n – 1)1 = 1 + n – 1 = n
\n2, 3, 4 ………………… are in A.P.
\nHere a = 2, d = 3 – 2 = 1
\ntn<\/sub> = a + (n – 1)d = 2 + (n – 1)1 = 2 + n – 1 = n + 1
\n3, 4, 5 …………….. are in A.P.
\nHere a = 3, d = 4 – 3 = 1
\ntn<\/sub> = a + (n – 1)d = 3 + (n – 1)1 = 3 + n – 1 = n + 2
\n\u2234 The nth term in the given series is n(n + 1) (n + 2).
\nLet S(n) be the statement that
\n1.2.3 + 2.3.4 + 3.4.5 + ……………….. + n (n + 1) (n + 2) = \\(\\frac{n(n+1)(n+2)(n+3)}{4}\\)
\nIf n = 1 then
\nL.H.S = n(n + 1) (n + 2) = 1(1 + 1) (1 + 2) = 1.2.3 = 6
\nR.H.S = \\(\\frac{n(n+1)(n+2)(n+3)}{4}\\) = \\(\\frac{1(1+1)(1+2)(1+3)}{4}\\) = \\(\\frac{1.2 .3 .4}{4}\\) = 6
\n\u2234 L.H.S = R.H.S
\n\u2234 S(1) is true Assume that s(k) is true
\n1.2.3 + 2.3.4 + 3.4.5 + ……………….. + k (k + 1) (k + 2) = \\(\\frac{\\mathrm{k}(\\mathrm{k}+1)(\\mathrm{k}+2)(\\mathrm{k}+3)}{4}\\)
\nAdding (k + 1) (k + 2) (k + 3) on both sides we get
\n1.2.3 + 2.3.4 + 3.4.5 + …………….. + k (k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)
\n\"TS
\n\u2234 S(k + 1) is true.
\n\u2234 By the principle of mathematical induction, S(n) is true \u2200 n \u2208 N.
\n\u2234 1.2.3 + 2.3.4 + 3.4.5 + ……………… + n(n + 1) (n + 2) = \\(\\frac{n(n+1)(n+2)(n+3)}{4}\\) \u2200 n \u2208 N.<\/p>\n

Question 9.
\nBy using mathematical induction show that \u2200 n \u2208 N,
\n\\(\\frac{1^3}{1}+\\frac{1^3+2^3}{1+3}+\\frac{1^3+2^3+3^3}{1+3+5}\\) + ……………….. upto n terms = \\(\\frac{n}{24}\\) [2n2<\/sup> + 9n + 13]. [Mar. 14, 07, 05]
\nAnswer:
\nNumerator: nth<\/sup> = 13<\/sup> + 23<\/sup> + 33<\/sup> + …………. + n3<\/sup> = \u03a3n3<\/sup> = \\(\\frac{n^2(n+1)^2}{4}\\)<\/p>\n

Denominator: 1 + 3 + 5 + ………………. are in A.P.
\nHere a = 1, d = 3 – 1 = 2
\ntn<\/sub> = a + (n – 1) d = 1 + (n – 1) 2 = 1 + 2n – 2 = 2n – 1
\n\u2234 nth<\/sup> term is
\n\"TS
\n\u2234 L.H.S = R.H.S
\n\u2234 S(1) is true.
\nAssume that S(k) is true.
\n\"TS<\/p>\n

Verification Method: \\(\\frac{\\mathrm{n}}{24}\\) [2n2<\/sup> + 9n + 13]
\nPut n = k + 1
\n= \\(\\frac{(\\mathrm{k}+1)}{24}\\) [2(k + 1)2<\/sup> + 9 (k + 1) + 13] = \\(\\frac{(\\mathrm{k}+1)}{24}\\) [2k2<\/sup> + 2 + 4k + 9k + 9 + 13]
\n= \\(\\frac{1}{24}\\) [2k3<\/sup> + 2k + 4k2<\/sup> + 9k2<\/sup> + 9k + 13k + 2k2<\/sup> + 2 + 4k + 9k + 9 + 13] = \\(\\frac{1}{24}\\) [2k3<\/sup> + 15k2<\/sup> + 37k + 24]
\n\u2234 S(k + 1) is true.
\nBy the principle of mathematical induction S(n) is true, \u2200 n \u2208 N.
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 10.
\nBy using mathematical induction show that \u2200 n \u2208 N, 12<\/sup> + (12<\/sup> + 22<\/sup>) + (12<\/sup> + 22<\/sup> + 32<\/sup>) + ……….. upto n terms = \\(\\frac{n(n+1)^2(n+2)}{12}\\).
\nAnswer:
\n\"TS<\/p>\n

Question 11.
\nBy using mathematical induction show that \u2200 n \u2208 N, 2 + 3.2 + 4.22<\/sup> + ……… upto n terms
\nAnswer:
\n2.1 + 3.2 + 4.22<\/sup> + ………………… upto n terms = n . 2n<\/sup>
\n2, 3, 4 ………….. are in A.P.
\nHere a = 2, d = 3 – 2 = 1
\ntn<\/sub> = a + (n – 1)d = 2 + (n – 1)1 = 2 + n – 1 = n + 1
\n1, 2, 22<\/sup>, ……………… are in G.P.
\nHere a = 1, r = \\(\\frac{2}{1}\\) = 2
\ntn<\/sub> = a . rn – 1<\/sup> = 1 . 2n – 1<\/sup> = 2n – 1<\/sup>
\n\u2234 The nth term in the given series is (n + 1) (2n – 1<\/sup>).
\nLet S(n) be the statement that
\n2.1 + 3.2 + 4.22<\/sup> + ……….. + (n + 1) 2n – 1<\/sup> = n . 2n<\/sup>
\nIf n = 1, then
\nL.H.S. = (n + 1)2n – 1<\/sup> = (1 + 1) 21 – 1<\/sup> = 2.20<\/sup> = 2.1 = 2
\nR.H.S. = n . 2n<\/sup> = 1.21<\/sup> = 2
\n\u2234 LHS = RHS
\n\u2234 S(1) is true.
\n[(n + 1)2n – 1<\/sup>
\nput n = k + 1
\n(k + 1 + 1) 2k + 1 – 1<\/sup>
\n(k + 2) . 2k<\/sup>]
\nAssume that S(k) is true.
\n2.1 + 3.2 + 4.22<\/sup> + ……….. + (k + 1)2k – 1<\/sup> = k . 2k
\nAdding (k + 2) . 2k<\/sup> on both sides, we get
\n2.1 + 3.2 + 4.22<\/sup> + ………… + (k + 1) 2k – 1<\/sup> + (k + 2) 2k<\/sup> = k. 2k<\/sup> + (k + 2) . 2k<\/sup> = 2k<\/sup> (k + k + 2)
\n= 2k<\/sup> (2k + 2) = 2k<\/sup> . 2 (k + 1) = (k + 1) . 2k + 1<\/sup>
\n\u2234 S(k + 1) is true.
\n\u2234 By the principle of mathematical induction, S(n) is true, \u2200 n \u2208 N.
\n2.1 + 3.2 + 4. 22<\/sup> + ……………… + (n + 1) 2n – 1<\/sup> = n . 2n<\/sup>, \u2200 n \u2208 N.<\/p>\n

\"TS<\/p>\n

Question 12.
\nBy using mathematical induction show that \u2200 n \u2208 N, 49n<\/sup> + 16n – 1 is divisible by 64 for all positive integer n. [Mar. 18 (TS); Mar. 17 (AP); May 13, 05, 98, 93]
\nAnswer:
\nLet S(n) be the statement that f(n) = 49n<\/sup> + 16n – 1 is divisible by 64.
\nIf n = 1, then
\nf(1) = 491<\/sup> + 16.1 – 1 = 49 + 16 – 1 = 49 + 15 = 64 = 64 \u00d7 1 is divisible by 64
\n\u2234 S(1) is true.
\nAssume that S(k) is true.
\nf(k) is divisible by 64 \u21d2 49k<\/sup> + 16k – 1 is divisible by 64
\n\u21d2 49k<\/sup> + 16k – 1 = 64 M for some integer M \u21d2 49k<\/sup> = 64M – 16k + 1
\nNow
\nf(k + 1) = 49k + 1<\/sup> + 16 (k + 1) – 1 = 49k<\/sup>. 49 + 16k + 16 – 1 = (64M – 16k + 1) 49 + 16k + 15
\n= 64.49M – 784k + 49 + 16k + 15 = 64.49M – 768k + 64
\n= 64(49M – 12k + 1) is divisible by 64. [ \u2235 49M – 12k + 1 is an integer]
\n\u2234 S(k + 1) is true.
\nBy the principle of mathematical induction S(n) is true \u2200 n \u2208 N.
\n\u2234 49n<\/sup> + 16n – 1 is divisible by 64, \u2200 n \u2208 N.<\/p>\n

Question 15.
\nBy using mathematical induction show that 3 \u2219 52n + 1<\/sup> + 23n + 1<\/sup> is divisible by 17 \u2200 n \u2208 N. [May 12, 10, 08, 01, ’96]
\nAnswer:
\nLet S(n) be the statement that f(n) = 3 . 52n + 1<\/sup> + 23n + 1<\/sup> is divisible by 17.
\nIf n = 1, then
\nf(1) = 3.52.1 + 1<\/sup> + 23.1 + 1<\/sup> = 3.53<\/sup> + 24 = 3(125) + 16 = 375 + 16 = 391 = 17 \u00d7 23 is divisible by 17
\n\u2234 S(1) is true.
\nAssume that S(k) is true.
\n\u2234 f(k) is divisible by 17.
\n\u21d2 3.52k + 1<\/sup> + 23k + 1<\/sup> is divisible by 17.
\n\u21d2 3.52k + 1<\/sup> + 23k + 1<\/sup> = 17 M for some integer M.
\n\u21d2 3.52k + 1<\/sup> = 17M – 23k + 1<\/sup>
\nNow f (k + 1) = 3 . 52(k + 1) + 1<\/sup> + 23(k + 1) + 1<\/sup> = 3.52k + 2 + 1<\/sup> + 23k + 3 + 1<\/sup>
\n= 3.52k + 1<\/sup> . 52<\/sup> + 23k + 1<\/sup> . 23<\/sup>
\n= (17M – 23k + 1<\/sup>) 25 + 8 . 23k + 1<\/sup> = 17.25 M – 25.23k + 1<\/sup> + 8 . 23k + 1<\/sup>
\n= 17.25 M – 17.23k + 1<\/sup>
\n= 17 (25 M – 23k + 1<\/sup>) is divisible by 17. [\u2235 25M – 23k + 1<\/sup> is an integer]
\n\u2234 S(k + 1) is true.
\nBy the principle of mathematical induction, S(n) is true \u2200 n \u2208 N.
\n\u2234 3 . 52n + 1<\/sup> + 23n + 1<\/sup> is divisible by 17, \u2200 n \u2208 N.<\/p>\n

Question 14.
\nBy using mathematical induction show that \u2200 n \u2208 N, xn<\/sup> – yn<\/sup> is divisible by x – y. [May 04]
\nAnswer:
\nLet S(n) be the statement that f(n) = xn<\/sup> – yn<\/sup> is divisible by (x – y).
\nIf n = 1 then
\nf(1) = x1<\/sup> – y1<\/sup> = x – y = (x – y) \u00d7 1 is divisible by (x – y)
\n\u2234 S(l) is true.
\nAssume that S(k) is true.
\nf(k) is divisible by (x – y).
\n\u21d2 xk<\/sup> – yk<\/sup> is divisible by (x – y)
\n\u21d2 xk<\/sup> – yk<\/sup> = (x – y) M, for some integer M.
\n\u21d2 xk<\/sup> = (x – y) M + yk<\/sup>
\nNow
\nf(k + 1) = xk + 1<\/sup> – yk + 1<\/sup> = xk<\/sup> . x – yk<\/sup> . y = [(x – y) M + yk<\/sup>] x – yk<\/sup> . y = (x – y) . M x + yk<\/sup> . x – yk<\/sup> . y
\n= (x – y) M x + yk<\/sup> (x – y) = (x – y) [Mx + yk<\/sup>] is divisible by (x – y).
\n[Mx + yk<\/sup>] is an integer.
\n\u2234 S(k + 1) is true.
\nBy the principle of mathematical induction S(n) is true, \u2200 n \u2208 N.
\n\u2234 xn<\/sup> – yn<\/sup> is divisible by (x – y), \u2200 n \u2208 N.<\/p>\n

\"TS<\/p>\n

Some More Maths 1A Mathematical Induction Important Questions<\/h3>\n

Question 1.
\nBy using mathematical induction show that \u2200 n \u2208 N, 43<\/sup> + 83<\/sup>+ 123<\/sup> + ……………. upto n terms = 16n2<\/sup> (n + 1)2<\/sup>.
\nAnswer:
\n4, 8, 12, are in A.P.
\nHere a = 4, d = t2<\/sub> – t1<\/sub> = 8 – 4 = 4
\ntn<\/sub> = a + (n – 1)d = 4 + (n – 1)4 = 4 + 4n – 4 = 4n
\n\u2234 The nth term in the given series is (4n)3<\/sup> = 64n3<\/sup>
\nLet S(n) be the statement that 43<\/sup> + 83<\/sup> + 123<\/sup> + …………….. + 64n3<\/sup> = 16n2<\/sup> (n + 1)2<\/sup>
\nIf n = 1, then
\nLHS = 64n3<\/sup> = 64(1)3<\/sup> = 64
\nRHS = 16n2<\/sup> (n + 1)2<\/sup> = 16.12<\/sup> (1 +!)2<\/sup> = 16.1.4 = 64
\n\u2234 LHS = RHS
\n\u2234 S(1) is true.
\nAssume that S(k) is true.
\n43<\/sup> + 83<\/sup> + 123<\/sup> + ………………….. + 64k3<\/sup> = 16k2<\/sup> (k + 1)2<\/sup>
\nAdding 64(k + 1)3<\/sup> on both sides, we get
\n43<\/sup> + 83<\/sup> + 123<\/sup> + …………….. + 64k3<\/sup> + 64(k + 1)3<\/sup> = 16k2<\/sup> (k + 1)2<\/sup> + 64 (k + 1)3<\/sup> = 16(k + 1)2<\/sup> [k2<\/sup> + 4 (k + 1)]
\n= 16(k + 1)2<\/sup> [k2<\/sup> + 4k + 4] = 16(k + 1)2<\/sup> (k + 2)2<\/sup> = 16(k + 1)2<\/sup> (k + 1 + 1)2<\/sup>
\n\u2234 S(k + 1) is true.
\nBy the principle of mathematical induction, S(n) is true \u2200 n \u2208 N.
\n\u2234 43<\/sup> + 83<\/sup> + 123<\/sup> + ………………. + 64n3<\/sup> = 16n2<\/sup> (n + 1)2<\/sup>, \u2200 n \u2208 N.<\/p>\n

Question 2.
\nBy using mathematical induction show that \u2200 n \u2208 N. 2.42n + 1<\/sup> + 33n + 1<\/sup> is divisible by 11.
\nAnswer:
\nLet S(n) be the statement that f(n) = 2.42n + 1<\/sup> + 33n + 1<\/sup> is divisible by 11.
\nIf n = 1, then
\nf(1) = 2.42.1 + 1<\/sup> + 33.1 + 1<\/sup> = 2.43<\/sup> + 34 = 2(64) + 81
\n= 128 + 81 = 209 = 11 \u00d7 19 is divisible by 11.
\n\u2234 S(1) is true.
\nAssume that S(k) is true.
\n\u2234 f(k) is divisible by 11.
\n\u21d2 2.42k + 1<\/sup> + 33k + 1<\/sup> is divisible by 11
\n\u21d2 2.42k + 1<\/sup> + 33k + 1<\/sup> = 11 M, for some integer M
\n\u21d2 2.42k + 1<\/sup> = 11 M – 33k + 1<\/sup>
\nNow f(k + 1) = 2.42(k + 1) + 1<\/sup> + 33(k + 1) + 1<\/sup> = 2.42k + 2 + 1<\/sup> + 33k + 3 + 1<\/sup> = 2.42k + 1<\/sup> . 42<\/sup> + 33k + 1<\/sup> . 33<\/sup>
\n= (11 M – 33k + 1<\/sup>) 16 + 27. 33k + 1<\/sup> = 11.16 M – 16. 33k + 1<\/sup> + 27.33k + 1<\/sup> = 11.16M + 11.33k + 1<\/sup>
\n= 11(16 M + 33k + 1<\/sup>) is divisible by 11 (\u2235 16 M + 33k + 1<\/sup> is an integer)
\n\u2234 S(k + 1) is true.
\nBy the principle of mathematical induction S(n) is true, \u2200 n \u2208 N.
\n\u2234 2.42n + 1<\/sup> + 33n + 1<\/sup> is divisible by 11, \u2200 n \u2208 N.<\/p>\n

\"TS<\/p>\n

Question 3.
\nUsing mathematical induction prove that statement for all \u2200 n \u2208 N,
\n\\(\\left(1+\\frac{3}{1}\\right)\\left(1+\\frac{5}{4}\\right)\\left(1+\\frac{7}{9}\\right) \\ldots\\left(1+\\frac{2 n+1}{n^2}\\right)\\) = (n + 1)2<\/sup>
\nAnswer:
\nLet S(n) be the statement that \\(\\left(1+\\frac{3}{1}\\right)\\left(1+\\frac{5}{4}\\right)\\left(1+\\frac{7}{9}\\right) \\ldots\\left(1+\\frac{2 n+1}{n^2}\\right)\\) = (n + 1)2<\/sup>
\nIf n = 1, then
\nLHS = 1 + \\(\\frac{2 n+1}{n^2}\\) = 1 + \\(\\frac{2.1+1}{(1)^2}\\) = 1 + \\(\\frac{2+1}{1}\\) = 1 + 3 = 4
\nRHS = (n + 1)2<\/sup> = (1 + 1)2<\/sup> = (2)2<\/sup> = 4
\n\u2234 LHS = RHS
\n\u2234 S(1) = is true
\nAssume that S(k) is true
\n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Mathematical Induction Important Questions to help strengthen their preparations for exams. TS Inter 1st Year Maths 1A Mathematical Induction Important Questions Question 1. By using mathematical induction show that \u2200 n \u2208 N, 12 + 22 + 32 + ……………… + … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[27],"tags":[],"yoast_head":"\nTS Inter 1st Year Maths 1A Mathematical Induction Important Questions - TS Board Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/tsboardsolutions.com\/maths-1a-mathematical-induction-important-questions\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"TS Inter 1st Year Maths 1A Mathematical Induction Important Questions - TS Board Solutions\" \/>\n<meta property=\"og:description\" content=\"Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Mathematical Induction Important Questions to help strengthen their preparations for exams. 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