Maths 1A Important Questions<\/a> TS Inter 1st Year Maths 1A Mathematical Induction Important Questions to help strengthen their preparations for exams.<\/p>\nTS 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
\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<\/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
\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>\nQuestion 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<\/p>\nVerification Method:
\n\\(\\frac{\\mathrm{n}\\left(\\mathrm{n}^2+6 n+11\\right)}{3}\\)
\nPut n = k + 1
\n
\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
<\/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<\/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<\/p>\n<\/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
\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>\nQuestion 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
\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<\/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
\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>\nQuestion 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>\nDenominator: 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
\n\u2234 L.H.S = R.H.S
\n\u2234 S(1) is true.
\nAssume that S(k) is true.
\n<\/p>\nVerification 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<\/p>\n<\/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<\/p>\nQuestion 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<\/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>\nQuestion 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>\nQuestion 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<\/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>\nQuestion 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> + 3