Question

Sum up to $16$ terms of the series $\frac{1^3}{1}+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\dots \dots$ is

• A. $850$
• B. $856$
• C. $816$
• D. $842$

Answer 1

The correct option is C $816$

$t_n=\frac{1^3+2^3+3^3+\dots +n^3}{1+2+3+\cdots +n}=\frac{{\left\{\frac{n(n+1)}{2}\right\}}^2}{\frac{n}{2}(n+1)}=\frac{n(n+1)}{2}=\frac{n^2}{2}+\frac{n}{2}$

$\therefore S_n=\sum t_n=\frac{1}{2}\sum n^2+\frac{1}{2}\sum n=\frac{1}{2}\times \frac{n(n+1)(2n+1)}{6}+\frac{1}{2}\times \frac{n(n+1)}{2}$

$\therefore S_{16}=\frac{16\cdot 17\cdot 33}{12}+\frac{16.17}{4}=816$