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156 | Mathematics (Class 11), Example 3. For all n > 1, prove that 1 n, , J Maeght ataad) med, 12 26 e, , Solution, We can write i hn, 1 saan, Bt eae, sate aD nel, 1, 12 ,, s true for some natural number #,, , We note that S(1), , Assume that S(k) ;, , 1 Fk, Le., Jt,i,1, aaa +, 12°23 3.4, , , , Mk+1) k+1, P(k + 1) is true whenever S(k) is true. We have, 1wiii 1 1, , . We need to Prove that, , 12°23 °34° "hea G+ Dh+2), , , , 1., k, ” hel Gah D (Usin, , 1 1 1 1 }- 1, “ Sata “hee |*e +Dikap, 1, , Ak+2)+1 — (h?+2k4+1)_ (kh +1)?, ~ (+ DR+2) = (ee Dik+2)~ (k+1)(k+2), k+l k+l, “h+2 eDel, - Hence, by the Principle of mathematical in, tion, S(n) is true for all natural numbers., Example 4-Prove that 2" > n for all positive integers n. Yn XY | (U. P.1!, Solution. Let S(n): 2" > n, When n = 1, 2! > 1. Hence P(1) is true., Assume that S(k) is true for any positive integer kh, i.e.,, sk, We shall now prove that S(k + 1) is true whe:, , never S(k) is true., Multiplying both sides of (1) by 2, we get, 2.24 > op, . . gke1, ie,, , >k=kiks k+1, Therefore, S(k + 1) is true when S(&) is try, , BEE es e. Hence, by Principle of mathematical ind, tion, S(7) is true for every positive.integer n,, Example 5. Prove that (1+ x" 2 (1+ nx), for all natural number n, where x > -1., Solution. Let S(n) be the given statement,, i S(n): (1 +x), on note that S() is true when n, , e, , 2(1+nx), forx>_4., = 1, since (1 +x) >(q +x)forxs_4
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| St) is true. Therefore, b, , | Positive intecar n, , v4, , ; hat Principle of Mathematical Induction |, 7 assume that,, , SR) (144 Peary hy) > —1 ig true,, We want to prove that S( 4. 1) is tr, , ' ‘i Ue for x > — 1 whenever S(h) is true, , Consider the identity ”, (ayer cy tx 4x), , Given that *>-Ieo(14x)0,, , Therefore, by using (14x) > ¢ , kx), we have, (4yyhea, , 214 hx\(1 4x), ies tas yyy hy + hy”),, Here kis a natural number and x?, , 2.0 60 that kx? > 0, Therefore, (tnt he 4 kx?), , . Plex 4 hx),, and so We obtain, , Le, 2[1+(1+h)x], , Thus, the statement in (2) is esta, , + gon, Slr) is true for all natural numb, ‘Example 6. For every positi, , ve integer n, Prove that 7" — 3" is divisible by 4., Solution. We can write, , ers,, , S(n) : 7? 3% ig divisible by 4., Wenote that S(1): 7-31 = 4 which is divisible by 4. Thus P(n) is true for n = 1,, Let S(k) be true for Some natural number k,, ie., Stk) : 7 — 3" is divisible by 4,, We can write 7* — 3 = 4d, where d ¢ N., , Now, we wish to Prove that S(k + 1) is true whenever S(k) is true,, Now THAD _ gh+1) = 74+) _ 7.38 47.3 _ gik+n, , = UT 3") + (7—3)3* = 1(4d) + (7 — 3) 3*, = T(4d) + 4.3* = 4(7d 4 34), From the last line, we see that 7+) _ 3+) is divisible by 4. Thus,, , S(k + 1) is true when, Yy principle of mathematical induction the state, , ment is tena fan .---—, 157, , vf), , wlZ), wfB), , blished. Hence, by the principle of mathematical induc
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160 | Mathematics (Class 11), , Example 11. Prove that for n ¢ Ny, , n, = —(n+1)(n4, nin + 3Xn + 8) = | (n+1Xn *OXn 49), , 1.4.7.4 2.6.8 4 wo, , Solution, Let the statement, , n, 4 6) = —(n+1K(n 4, Sn) 51.4.7 4 2.6.8 + ae FC 3yin +6) = 4 *Bxn 4,, , Porn = 1,, nf +7), ‘in tan ween +, S(1): 1.4.7 4, 1, = S(1): 28 = read, > S(1) : 28 = 28., “. S(1) is true., , Now we assume that for n =r ¢ N, the statement, , :, S(r) 114.74 2.6.8 4 ane tr + Br + 6) = GO FD + 6Nr +7), , is true. Then,, , 1.4.7 4.2.5.8 + sou. tr(r + Br +6) + (r+ 1) (r+ 4)(r +7), , = - (r+ Ir + Ort 7) + (r+ Der + 44) /, , =(r+1)(r+7) [Fr+o+0+4], , , , CD rTM? +6r+4r +16), , , , = CD ranir? +10r +16), , -o», , , , (r+7)[(r + 2)\(r + 8)), , zi 0, +2\r+7)(7 +8), , , , => ap + 1)is true., ; = S(r*) is true., Hence by the principle of mathematical induction, the given statement is true foral:, Pre, , Example 12. Prove that, , Solution. Let S(n) be the given statement,, , be, Sin)? 4224 an, 2, We note that S(n) is true for n = 1 since 125 =
