Question

What is the sum of $11+22+33+...n^2$?

• A. $\frac{n(n-1)(2n+1)}{6}$
• B. $\frac{n(n+1)(n+1)}{6}$
• C. $\frac{n(n+1)(2n+1)}{6}$
• D. $\frac{n(n+1)(2n+1)}{3}$

Answer 1

The correct option is C $\frac{n(n+1)(2n+1)}{6}$

Here $S_n=11+22+33+...n^2$
We consider the identity $x^3-(x-1{)}^3=3x^2-3x+1$
When x = 1, $1^3-(1-1{)}^3=3(1{)}^2-3\left(1\right)+1$
When x = 2, $2^3-(2-1{)}^3=3(2{)}^2-3\left(2\right)+1$
When x = 3, $3^3-(3-1{)}^3=3(3{)}^2-3\left(3\right)+1$.......
$n^3-(n-1{)}^3=3n^2-3n+1$
Adding vertically both sides,
$n^3=3(1^2+2^2+3^2+...n^2)-3(1+2+\mathrm{3...}+n)+(1+1+\mathrm{1..}n)$
$3(1^2+2^2+3^2+...n^2)=n^3+\frac{3n(n+1)}{2}-n$
= $\frac{2n^3+3n^2+n}{2}$
= $\frac{n(2n^2+3n+1)}{2}$
$\sum n^2=\frac{n(n+1)(2n+1)}{6}$