Question

Find \[\displaystyle\sum_{r=1}^{n}\frac{1^3 + 2^3 + 3^3 +...+r^3}{(r + 1)^2}\]

Answer 1

\[\displaystyle\sum_{r=1}^{n}\frac{1^3 + 2^3 + 3^3 +...+r^3}{(r + 1)^2}\]

\[\displaystyle\sum_{r=1}^{n}\frac{r^2(r+1)^2}{4} \times \frac{1}{(r+1)^2}\]

= \[\frac{1}{4}\displaystyle\sum_{r=1}^{n}r^2\]

= `1/4*("n"("n" + 1)(2"n" + 1))/6`

= `("n"("n" + 1)(2"n" + 1))/24`.