Question

$1^2+(1^2+2^2)+(1^2+2^2+3^2)+...+n$ brackets $=$

• A. $\frac{n(n+1{)}^2(n+2{)}^2}{12}$
• B. $\frac{n(n+1{)}^2(n+2)}{12}$
• C. $\frac{n^2(n+1)(n+2)}{12}$
• D. $\frac{(n+1)}{2}$

Answer 1

The correct option is B $\frac{n(n+1{)}^2(n+2)}{12}$

The following series is:

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

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

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).

$\therefore S_n=\frac{n^2(n+1{)}^2}{12}+\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{12}$

$=\frac{n(n+1)}{12}\times n^2+3n+2$

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