Question

Prove the following by using Mathematical induction :

1^2 + 2^2 + 3^2+ ..........n^2 =
[n (n+1)(2n+1)]/6 ​

Answer 1

SOLUTION

Prove that 12 + 22 + ... + n2 = [n(n+1)(2n+1)]/6, for every positive integer n.

Proof:

I. Prove that the equation holds for n = 1

If n = 1, then [n(n+1)(2n+1)]/6 = [(1)(2)(3)]/6 = 1.

So, 12 = [(1)(1+1)(2(1)+1)]/6

II. Assume that the equation holds for n and prove that the equation holds for n+1.

Assume: 12 + 22 + ... + n2 = [n(n+1)(2n+1)]/6

Prove: 12 + 22 + ... + n2 +(n+1)2 =

[(n+1)(n+2)(2n+3)]/6

By assumption, 12 + 22 +... + n2 + (n+1)2 =

[n(n+1)(2n+1)]/6 + (n+1)2

= [n(n+1)(2n+1) +6(n+1)2]/6

= [(n+1){n(2n+1) + 6(n+1)}]/6

=[(n+1){2n2 + 7n +6}]/6

= [(n+1)(n+2)(2n+3)]/6

Therefore, by induction on n, the equation is valid for all positive integers, n.

HOPE IT HELPS

Answer 2

Refers to the attachment

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