Sum of series $\sum^{10}_{i=1}i\bigg(\frac{1^2}{1+i}+\frac{2^2}{2+i}+\cdots \cdots +\frac{10^2}{10+i}\bigg)$

Question

The Sum of series $$\sum^{10}_{i=1}i\bigg(\frac{1^2}{1+i}+\frac{2^2}{2+i}+\cdots \cdots +\frac{10^2}{10+i}\bigg)$$

Try: Let $$S=\sum^{10}_{i=1}\frac{i}{1+i}+2^2\sum^{10}_{i=1}\frac{i}{2+i}+\cdots \cdots \cdots +10^2\sum^{10}_{i=1}\frac{i}{10+i}$$

$$S=\sum^{10}_{i=1}\sum^{10}_{j=1}\bigg[\frac{i}{i+j}-1\bigg]-100$$

Could some help me How to solve it, Thanks in advanced

Answer 1

The sum to calculate is $$ S=\sum_{i,j=1}^{10}\frac{j^2i}{i+j}=[\text{rename }i\leftrightarrow j]=\sum_{i,j=1}^{10}\frac{i^2j}{j+i}. $$ Hence $$ 2S=\sum_{i,j=1}^{10}\frac{i^2j+j^2i}{i+j}=\sum_{i,j=1}^{10}\frac{ij(i+j)}{i+j}=\sum_{i,j=1}^{10}ij=\Big(\sum_{i=1}^{10}i\Big)\Big(\sum_{j=1}^{10}j\Big)=\Big(\sum_{k=1}^{10}k\Big)^2=55^2. $$ Finally $$ S=\frac{55^2}{2}=\frac{(50+5)^2}{2}=\frac{2500+500+25}{2}=\frac{3025}{2}. $$

Answer 2

HINT:

Write $$\begin{align}S&=\sum^{10}_{i=1}\frac{i}{1+i}+2^2\sum^{10}_{i=1}\frac{i}{2+i}+\cdots+10^2\sum^{10}_{i=1}\frac{i}{10+i}\\&=\sum^{10}_{i=1}\frac{i}{1+i}+4\sum^{11}_{i=2}\left(\frac{i}{1+i}-\frac1{1+i}\right)+\cdots+100\sum^{19}_{i=10}\left(\frac{i}{1+i}-\frac9{1+i}\right)\end{align}$$ and combine like expressions.