Question

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

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

Answer 1

Answer: (D)

Here, $S_n=1^2+2^2+3^2+\dots +n^2$
Consider the identity, $K^3-(K-1{)}^3=3K^2-3K+1$
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
$n^3-0^3=3\left(1^2+2^2+3^2+4^2+\dots +n^2\right)-3(1+2+\dots +n)+n$
$n^3=3\sum _{k=1}^n{K}^2-3\sum _{k=1}^{n}K+n$
Now, $S_n=\sum _{K=1}^n{K}^2=\frac{1}{3}\left[n^3+3\sum _{K=1}^{n}K-n\right]$
$=\frac{1}{3}\left[n^3+\frac{3n(n+1)}{2}-n\right]$
$\left[\because \sum _{k=1}^{n}K=1+2+3+\dots +n=\frac{n(n+1)}{2}\right]$
$=\frac{1}{6}\left(2n^3+3n^2+3n-2n\right)$
$=\frac{1}{6}\left(2n^3+3n^2+n\right)=\frac{n(n+1)(2n+1)}{6}$