Evaluate: $\sum_{n=1}^{10}n\cdot (\sum_{r=1}^{10}\frac{r^2}{r+n})$

Question

Evaluate: $\sum_{n=1}^{10}n\cdot (\sum_{r=1}^{10}\frac{r^2}{r+n})$

using the property of sigma operator I put $n$ in second sigma

So it becomes: $\sum_{n=1}^{10}\cdot (\sum_{r=1}^{10}\frac{r^2n}{r+n})$

Which property I need to use now?

Answer 1

$S= \displaystyle \sum_{n=1}^{10} \left(\sum_{r=1}^{10}\dfrac{r^2n}{r+n}\right) = \sum_{n=1}^{10} \left(\sum_{r=1}^{10}\dfrac{n^2r}{r+n}\right)$

So:
$2S =\displaystyle \sum_{n=1}^{10} \left(\sum_{r=1}^{10}\dfrac{r^2n+n^2r}{r+n}\right) = \displaystyle \sum_{n=1}^{10} \left(\sum_{r=1}^{10} nr\right) \left( \dfrac{10\times 11}{2}\right)^2 = 3025$