Prove by principle of Mathematical induction that sum of squares of first n natural
number is {n(n+1)(2n+1)}/6
1
Expert's answer
2012-05-11T08:42:38-0400
Let n=1 Then {n(n+1)(2n+1)}/6 = 1*2*3/6 = 1 and
1^2=1 as well.
Suppose that we have proved that 1^2 + 2^2 + ... + k^2 = {k(k+1)(2k+1)}/6 for all k<n+1 Let us prove this identity for k=n+1, that is 1^2 + 2^2 + ... + n^2 + (n+1)^2 = {(n+1)(n+2)(2(n+1)+1)}/6
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