1. Chapter 4 Class 11 Mathematical Induction
2. Serial order wise

Transcript

Ex 4.1,2: Prove the following by using the principle of mathematical induction 13 + 23 + 33+โฆโฆ+ n3 = (๐(๐+1)/2)^2 Let P (n) : 13 + 23 + 33 + 43 + โฆ..+ n3 = (๐(๐+1)/2)^2 For n = 1, L.H.S = 13 = 1 R.H.S = (1(1 + 1)/2)^2= ((1 ร 2)/2)^2= (1)2 = 1 Hence, L.H.S. = R.H.S โด P(n) is true for n = 1 Assume that P(k) is true 13 + 23 + 33 + 43 + โฆ..+ k3 = (๐(๐ + 1)/2)^2 We will prove that P(k + 1) is true. 13 + 23 + 33 + โฆ..+ k3 + (k + 1)3= ((k + 1)((k + 1)+ 1)/2)^2 13 + 23 + 33 + โฆ..+ k3 + (k + 1)3= ((k + 1)(๐ + 2)/2)^2 We have to prove P(k+1) from P(k) i.e. (2) from (1) From (1) 13 + 23 + 33 + 43 + โฆ..+ k3 = (๐(๐ + 1)/2)^2 Adding (k+1)3 both sides, 13 + 23 + 33 + โฆ..+ k3 + (k + 1)3 = (๐(๐ + 1)/2)^2+ (k + 1)3 = (๐^2 (๐ + 1)^2)/2^2 + (k + 1)3 = (๐^2 (๐ + 1)^2)/4 + (k + 1)3 = (k^2 (k + 1)^2 + 4(๐ + 1)^3)/4 = ((k + 1)^2 (k^2 + 4๐ + 4))/4 = ((k + 1)^2 (k^2 + 2๐ + 2๐ + 4))/4 = ((k + 1)^2 (๐(๐ + 2) + 2(๐ + 2)))/4 = ((k + 1)^2 ((๐ + 2) (๐ + 2)))/4 = ((k + 1)^2 (k + 2)^2)/4 = ((k + 1)(๐ + 2)/2)^2 Thus, 13 + 23 + 33 + โฆ..+ k3 + (k + 1)3= ((k + 1)(๐ + 2)/2)^2 i.e. P(k + 1) is true whenever P(k) is true โดBy the principle of mathematical induction, P(n) is true for n, where n is a natural number

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