Question

Sum of squares of the first $n$ natural numbers i.e., $1^2+2^2+3^2+4^2+\ldots \ldots+n^2$ is

• A. $S_n=\frac{n(n-1)(2 n-1)}{6}$
• B. $S_n=\frac{n(n+1)(2 n+1)}{4}$
• C. $S_n=\frac{n(n+1)(2 n-1)}{4}$
• D. $S_n=\frac{n(n+1)(2 n+1)}{6}$

Answer 1

Answer: (D)

Here, $S_n=1^2+2^2+3^2+\ldots+n^2$
Consider the identity, $K^3-(K-1)^3=3 K^2-3 K+1$
Putting $K=1,2, \cdots, n$ successively, we obtain
$ 1^3-0^3=3(1)^2-3(1)+1 $
$ 2^3-1^3=3(2)^2-3(2)+1 $
$ 3^3-2^3=3(3)^2-3(3)+1$
$........................$
$........................$
$.........................$
$n^3-(n-1)^3=3(n)^2-3(n)+1$
Adding both sides, we get
$ n^3-0^3=3\left(1^2+2^2+3^2+4^2+\ldots+n^2\right)-3(1+2+\ldots+n)+n$
$ n^3=3 \displaystyle\sum_{k=1}^n K^2-3 \displaystyle\sum_{k=1}^n K+n $
Now, $ S_n=\displaystyle\sum_{K=1}^n K^2=\frac{1}{3}\left[n^3+3 \displaystyle\sum_{K=1}^n K-n\right] $
$ =\frac{1}{3}\left[n^3+\frac{3 n(n+1)}{2}-n\right] $
$ {\left[\because \displaystyle\sum_{k=1}^n K=1+2+3+\ldots+n=\frac{n(n+1)}{2}\right]} $
$ =\frac{1}{6}\left(2 n^3+3 n^2+3 n-2 n\right) $
$=\frac{1}{6}\left(2 n^3+3 n^2+n\right)=\frac{n(n+1)(2 n+1)}{6}$