Chapter 4 Class 11 Mathematical Induction
Concept wise

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Ex 4.1, 1 Prove the following by using the principle of mathematical induction for all n β N: 1 + 3 + 32+β¦β¦+ 3n β 1 = ((3π β 1))/2 Let P(n) : 1 + 3 + 32+β¦β¦+ 3n β 1 = ((3π β 1))/2 Proving for n = 1 For n = 1, L.H.S = 1 R.H.S = ((3^1 β 1))/2 = ((3 β 1))/2 = ((2))/2 = 1 Since, L.H.S. = R.H.S β΄ P(n) is true for n = 1 Proving P(k + 1) is true if P(k) is true Assume that P(k) is true, P(k): 1 + 3 + 32 +β¦..+ 3k β 1 = ((3π β 1))/2 We will prove that P(k + 1) is true. P(k + 1): 1 + 3 + 32 +β¦..+ 3(k + 1) β 1 = ((3^(π+1) β 1))/2 P(k + 1): 1 + 3 + 32 +β¦..3(k β 1) + 3(k) = ((3^(π+1) β 1))/2 We have to prove P(k + 1) is true Solving LHS 1 + 3 + 32 +β¦..+ 3k β 1 + 3k From (1): 1 + 3 + 32 +β¦..+ 3k β 1 = ((ππ β π))/π = ((ππ β π))/π + 3k = ((3π β 1) + 2 Γ 3^π)/2 = (ππ + π Γ π^π β π)/π = ( 3(3^π )β 1)/2 = (π^(π + π) β π)/π = RHS β΄ P(k + 1) is true when P(k) is true Thus, By the principle of mathematical induction, P(n) is true for n, where n is a natural number