Question

Find the sum of the squares of the first n natural numbers.

A. $\frac{n^2(n+1)}{6}$

B. $\frac{n(n+1)(n-1)}{6}$

C. $\frac{n(2n+1)}{6}$

D. $\frac{n(n+1)(2n+1)}{6}$

Answer 1

The correct option is D.

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

Given series $s$ = $1^2$ + $2^2$ + $3^2$ + $4^2$ +.................... +$n^2$

We know,

$n^3$ - $(n-1{)}^3$ = 3$n^2$ - $3n+1$

$(n-1{)}^3$ - $(n-2{)}^3$ = 3$(n-1{)}^2$ - 3$(n-1)$ + 1

$(n-2{)}^3$ - $(n-3{)}^3$ = 3$(n-2{)}^2$ - 3$(n-2)$ + 1

Adding both LHS and RHS

$n^3=3{\sum }_{i=1}^n{n}^2-3{\sum }_{i=1}^{n}n+n$

$n^3=3{\sum }_{i=1}^n{n}^2-3\frac{\left(n\right)(n+1)}{2}+n$

$n^3=3{\sum }_{i=1}^n{n}^2-\frac{(3n^2+3n)+2n}{2}$

$n^3=3{\sum }_{i=1}^n{n}^2-\frac{3n^2-3n+2n}{2}$

$n^3=3{\sum }_{i=1}^n{n}^2-\frac{3n^2+n}{2}$

${\sum }_{i=1}^n{n}^2=\frac{(2n^3+3n^2+n)}{6}=\frac{n(2n^2+3n+1)}{6}$

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