Question

Let $S=3+\frac{5(1^3+2^3)}{1^2+2^3}+\frac{7(1^3+2^3+3^3}{1^2+2^2+3^2}+...$ Then the sum up to 10 terms is

• A. $220$
• B. $660$
• C. $330$
• D. $1320$

Answer 1

The correct option is B $660$

General term of given series $\left(T_n\right)$
$T_n=\frac{(2n+1)(1^3+2^3+3^3...n)}{(1^2+2^2+....n)}$
$T_n=\frac{(2n+1){\left(\frac{\left(n\right)(n+1)}{2}\right)}^2}{\frac{n(n+1)(2n+1)}{6}}$
$T_n=\frac{3}{2}n(n+1)\Rightarrow T_n=\frac{3}{2}[n^2+n]$
sum of series $={\sum }_{n=1}^{10}{T}_n=\frac{3}{2}[\frac{10\left(11\right)\left(21\right)}{6}+\frac{10.11}{2}]\frac{3}{2}\left[440\right]=660$