Question

Compute
\[\frac{1^2}{1^2-10+50}+\frac{2^2}{2^2-20+50}+\frac{3^2}{3^2-30+50}+\cdots +\frac{9^2}{9^2-90+50}\]

Answer 1

\[(n)^{2}/[(n)^{2}-(n \times10)+50]\]
where n is no of tern

Answer 2

term*

Answer 3

did u get it ?
and was it right???

Answer 4

yes of course, so what? the problem wants us to compute the sum in a smart way.

Answer 5

wht do u mean

Answer 6

nvm, what is the sum of all the above numbers?

Answer 7

where does it end???

Answer 8

until n=9.

Answer 9

The above sum may be represented as \[a=\sum_{n=1}^{9}\frac{n^2}{n^2-10n+50}\] Now, \[a=1+\sum_{n=1}^{4}\frac{n^2+(10-n)^2}{n^2-10n+50}=1+2\sum_{n=1}^{4}\frac{n^2-10n+50}{n^2-10n+50}=1+2\cdot4=9\].