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 $s=1^2+2^2+3^2+4^2+\cdots +n^2$

We know,
$n^3-(n-1{)}^3=3n^2-3n+1$
$(n-1{)}^3-(n-2{)}^3=3\left(\right(n-1{)}^2-3(n-1)+1$
$(n-2{)}^3-(n-3{)}^3=3(n-2{)}^2-3(n-2)+1$
. .
. .
On adding both L.H.S and R.H.S, we get
$n^3=3{\sum }_{i=1}^n{n}^2-3{\sum }_{i=1}^{n}n+n$

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

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

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

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

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

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