Mathematics - Sequence & Series Question with Solution | TestHub

The $k$ th numerator is $\left(1^3+\cdots +k^3\right)$ and the denominator is $k^2$, so the term is $(k+1{)}^2/4$. Hence the sum is $\frac{1}{4}{\sum }_{k=1}^{16}(k+1{)}^2=\frac{1}{4}[(17\cdot 18\cdot 35)/6-1]=446$.