Question

$1^2+(1^2+2^2)+(1^2+2^2+3^2)+...$ to $n$ terms.

• A. $\frac{1}{12}n(n+1)(2n^3+3n^2+n)$
• B. $\frac{1}{12}n(n+1)(3n^3+2n^2+n)$
• C. $\frac{1}{6}n(n+1)(2n+1)$
• D. $\frac{1}{4}{n}^2(n+1{)}^2$

Answer 1

The correct option is A $\frac{1}{12}n(n+1)(2n^3+3n^2+n)$

nth term $T_n=1^2+2^2+3^2+.....+n^2=\sum n^2=\frac{1}{6}n(n+1)(2n+1)=\frac{1}{6}(2n^3+3n^2+n)$

Thus required summation is $=\sum T_n=\frac{1}{6}(\sum 2n^3+\sum 3n^2+\sum n)$

$=\frac{1}{6}(2\frac{n^2(n+1{)}^2}{4}+3\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2})=\frac{1}{12}n(n+1)(2n^3+3n^2+n)$