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Question
Find \[\displaystyle\sum_{r=1}^{n}\frac{1^3 + 2^3 + 3^3 +...+r^3}{(r + 1)^2}\]
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Solution
\[\displaystyle\sum_{r=1}^{n}\frac{1^3 + 2^3 + 3^3 +...+r^3}{(r + 1)^2}\]
\[\displaystyle\sum_{r=1}^{n}\frac{r^2(r+1)^2}{4} \times \frac{1}{(r+1)^2}\]
= \[\frac{1}{4}\displaystyle\sum_{r=1}^{n}r^2\]
= `1/4*("n"("n" + 1)(2"n" + 1))/6`
= `("n"("n" + 1)(2"n" + 1))/24`.
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