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

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

## closed as off-topic by Carl Mummert, Saad, Namaste, Claude Leibovici, cansomeonehelpmeoutMay 13 '18 at 10:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Saad, Namaste, Claude Leibovici, cansomeonehelpmeout
If this question can be reworded to fit the rules in the help center, please edit the question.

• My 1st attempt was to compute this sum via Maxima: sum(i*sum(j^2/(j+i),j,1,10),i,1,10); It is 3025/2. Your final expression for S is wrong: sum(sum(i/(i+j)-1,j,1,10),i,1,10)-100; gives us -150. – szw1710 May 9 '18 at 8:07
• I've voted to close this question because there's no genuine effort at all. Also, the OP keeps posting questions with only superficial tries to circumvent being closed as off-topic. – Saad May 11 '18 at 23:45

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 $$S=\frac{55^2}{2}=\frac{(50+5)^2}{2}=\frac{2500+500+25}{2}=\frac{3025}{2}.$$
• Note the interesting trick: $$\color{red}5\color{blue}5^2=\overbrace{\color{red}{5\times(5+1)}\color{blue}{25}}^{\text{concatenate}}=3025\quad!$$ – TheSimpliFire May 12 '18 at 7:44
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.
• You have the like terms $\dfrac i{1+i}$ which are summed over $1-10$, $2-11$, etc. So just add them together, taking note of the squares $4,9,\cdots$ in front. Do the same for $\dfrac1{1+i},\cdots,\dfrac9{1+i}$, noting that the $1,\cdots,9$ can be taken out the summations. – TheSimpliFire May 9 '18 at 11:08