$$1^2 + 2^2 + 3^2 + … + n^2 = \dfrac{[n (n+1) (2n+1)]}{6}$$

Asked by Aaryan | 1 year ago |  77

##### Solution :-

For, n = 1

P (1) = $$\dfrac{[1 (1+1) (2+1)]}{6}$$

1 = 1

P (n) is true for n = 1

Let P (n) is true for n = k, so

P (k): 12 + 22 + 32 + … + k2 =$$\dfrac{ [k (k+1) (2k+1)]}{6}$$

Let’s check for P (n) = k + 1, so

P (k) = 12 + 22 + 32 + – – – – – + k2 + (k + 1)2 = $$\dfrac{ [k + 1 (k+2) (2k+3)] }{6}$$

= 12 + 22 + 32 + – – – – – + k2 + (k + 1)2

= $$\dfrac{ (k +1) [2k^2 + 7k + 6]}{6}$$

= $$\dfrac{ (k +1) [2k^2 + 4k + 3k + 6]}{6}$$

=$$\dfrac{ [(k +1) (2k + 3) (k + 2)] }{6}$$

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

Answered by Aaryan | 1 year ago

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