Question

Find the sum of the series

$\left(2^2\right)+(2^2+4^2)+(2^2+4^2+6^2)+\dots (2^2+4^2+6^2+\dots \cdots +(2n{)}^2\left)\right(nterms)$

• A. $\frac{n(n+2{)}^2(n+1)}{3}$
• B. $\frac{n(n+1{)}^2(n+2)}{3}$
• C. $\frac{\left\{n\right(n+1\left)\right(n+2){\}}^2}{3}$
• D. $\frac{n(n+1)(n+2)}{3}$

Answer 1

The correct option is B $\frac{n(n+1{)}^2(n+2)}{3}$

Take $2^2$ common out of the whole series

What is left is $\left(1^2\right)+(1^2+2^2)+...$ n terms

The general form of each term is $\frac{\left(k\right)(k+1)(2k+1)}{6}$

The answer we want to calculate is

$2^2{\sum }_{k=1}^n\frac{\left(k\right)(k+1)(2k+1)}{6}$

$S_n=2^2{\sum }_{k=1}^n\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$

We know

${\sum }_{i=1}^{n}k=1+2+3+\dots +n=\frac{n(n+1)}{2}$ (sum of first n natural numbers)

${\sum }_{i=1}^n{k}^2=1^2+2^2+3^2+\dots +n^2=\frac{n(n+1)(2n+1)}{6}$ (sum of squares of the first n natural numbers)

${\sum }_{i=1}^n{k}^3=1^3+2^3+3^3+\dots +n^3=\frac{n^2(n+1{)}^2}{12}$ (sum of cubes of the first n natural numbers).

$\therefore S_n=2^2(\frac{n^2{(n+1)}^2}{4\cdot 3}+\frac{n(n+1)(2n+1)}{6\cdot 2}+\frac{n(n+1)}{2\cdot 6})$

$=4\frac{{\left(n\right)}^2{(n+1)}^2(n+2)}{12}$

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

Hence. option 'B' is correct.