solve it
Q. Find
(i) $\sum _{r=0}^9\frac{r^2}{r^2+{\left(9-r\right)}^2}$

Dear student
As to solve this we need higher level concepts. So, solving it by the method which is easily understandable to you.

$$\begin{gathered} \sum _{\mathrm{r}=0}^9\frac{{\mathrm{r}}^2}{{\mathrm{r}}^2+(9-\mathrm{r}{)}^2} \\ =0+\frac{1^2}{1^2+(9-1{)}^2}+\frac{2^2}{2^2+(9-2{)}^2}+\frac{3^2}{3^2+(9-3{)}^2}+\frac{4^2}{4^2+(9-4{)}^2}+\frac{5^2}{5^2+(9-5{)}^2}+\frac{6^2}{6^2+(9-6{)}^2}+\frac{7^2}{7^2+(9-7{)}^2}+\frac{8^2}{8^2+(9-8{)}^2}+\frac{9^2}{9^2+(9-9{)}^2} \\ =0+\frac{1}{1+8^2}+\frac{4}{4+7^2}+\frac{9}{9+6^2}+\frac{16}{16+5^2}+\frac{25}{25+4^2}+\frac{36}{36+3^2}+\frac{49}{49+2^2}+\frac{64}{64+1^2}+\frac{81}{81+0} \\ =\frac{1}{65}+\frac{4}{53}+\frac{1}{5}+\frac{16}{41}+\frac{25}{41}+\frac{4}{5}+\frac{49}{53}+\frac{64}{65}+1 \\ =\left(\frac{1}{65}+\frac{64}{65}\right)+\left(\frac{4}{53}+\frac{49}{53}\right)+\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{16}{41}+\frac{25}{41}\right)+1 \\ =1+1+1+1+1 \\ =5 \end{gathered}$$
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