Question

The sum of first 12 terms of the series $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+........$ is

• A. 204.5
• B. 126.25
• C. 216.5
• D. 156.25

Answer 1

The correct option is A 204.5

Let $T_n$ be the nth term of the given series. Then, $T_n=\frac{1^3+2^3+3^3+.....+n^3}{1+3+5+...+(2n-1)}\Rightarrow T_n=\frac{{\left(\frac{n\left(n+1\right)}{2}\right\}}^2}{\frac{n}{2}\left(1+(2n-1)\right\}}=\frac{\frac{n^2{\left(n+1\right)}^2}{4}}{\frac{n}{2}\times 2n}=\frac{{\left(n+1\right)}^2}{4}\Rightarrow T_n=\frac{n^2+2n+1}{4}Thesumoffirstntermswillbegivenby{S}_n={\sum }_{k=1}^n{T}_k={\sum }_{k=1}^n\left(\frac{n^2+2n+1}{4}\right)\Rightarrow S_n=\frac{1}{4}\left({\sum }_{k=1}^n{n}^2+2{\sum }_{k=1}^{n}n+{\sum }_{k=1}^{n}1\right]\Rightarrow S_n=\frac{1}{4}\left(\frac{n(n+1)(2n+1)}{6}+2\frac{n(n+1)}{2}+n\right]\Rightarrow S_n=\frac{n}{24}\left(2n^2+9n+13\right]Forn=12S_n=\frac{12}{24}\left(2{\left(12\right)}^2+9\left(12\right)+13\right]=204.5$