Summation-1

Similar solution as @Alak Bhattacharya 's

$\begin{aligned} S & = \sum_{n=1}^9 \frac {n^2}{n^2-10n+50} & \small \blue{\text{By reflection }\sum_{k=a}^b f(k) = \sum_{k=a}^b f(a+b-k)} \\ & = \frac 12 \sum_{n=1}^9 \left(\frac {n^2}{n^2-10n+50} + \frac {(10-n)^2}{(10-n)^2-10(10-n)+50}\right) \\ & = \frac 12 \sum_{n=1}^9 \left(\frac {n^2}{n^2-10n+50} + \frac {n^2-20n+100}{n^2-10n+50}\right) \\ & = \frac 12 \sum_{n=1}^9 \frac {2n^2-20n+100}{n^2-10n+50} \\ & = \frac 12 \sum_{n=1}^9 2 = \sum_{n=1}^9 1 = \boxed 9 \end{aligned}$

The $n^{th}$ term is $t_n=\dfrac{n^2}{n^2-10n+50}$ . So, $t_{10-n}=\dfrac{(10-n)^2}{(10-n)^2-10(10-n)+50}=\dfrac{(10-n)^2}{n^2-10n+50}$ . Therefore $t_n+t_{10-n}=2$ for $n\neq 10-n$ or $n\neq 5$ and $t_5=1$ . Hence the required sum is $4\times 2+1=\boxed 9$ .