Question

The sum of the series $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots \dots$ upto $16$ terms is

• A. $246$
• B. $646$
• C. $446$
• D. $746$

Answer 1

The correct option is C $446$

Given : $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots \dots$
Now,
$T_n=\frac{\sum _{r=1}^n{r}^3}{\sum _{r=1}^n(2r-1)}\Rightarrow T_n=\frac{1}{4}\times \frac{n^2(n+1{)}^2}{2\left(\frac{n(n+1)}{2}\right)-n}\Rightarrow T_n=\frac{1}{4}(n+1{)}^2\Rightarrow S_{16}=\frac{1}{4}(2^2+3^2+4^2+\cdots {17}^2)\Rightarrow S_{16}=\frac{1}{4}(1^2+2^2+3^2+4^2+\cdots {17}^2-1^2)\Rightarrow S_{16}=\frac{1}{4}(\frac{17(17+1)(2\cdot 17+1)}{6}-1)\therefore S_{16}=446$