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


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


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

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  • $\begingroup$ 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. $\endgroup$ – szw1710 May 9 '18 at 8:07
  • $\begingroup$ 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. $\endgroup$ – 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 $$ S=\frac{55^2}{2}=\frac{(50+5)^2}{2}=\frac{2500+500+25}{2}=\frac{3025}{2}. $$

  • 1
    $\begingroup$ Note the interesting trick: $$\color{red}5\color{blue}5^2=\overbrace{\color{red}{5\times(5+1)}\color{blue}{25}}^{\text{concatenate}}=3025\quad!$$ $\endgroup$ – 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.

  • $\begingroup$ The SimpliFire plz explain me in detail. tThanks $\endgroup$ – DXT May 9 '18 at 11:03
  • $\begingroup$ 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. $\endgroup$ – TheSimpliFire May 9 '18 at 11:08

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