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Sum of Squares of First N Natural Numbers

The sum of th...

Question

The sum of the squares of the first n natural numbers is

\frac{n (n + 1)}{2}

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\frac{n (n + 1) (2 n + 1)}{6}

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\frac{n (n + 1) (n + 2)}{12}

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\frac{n (n + 1) (2 n + 1)}{12}

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Solution

The correct option is B $\frac{n (n + 1) (2 n + 1)}{6}$
The formula to find sum of first $n$ terms of a sequence of squares of natural numbers $= \frac{n (n + 1) (2 n + 1)}{6}$

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