Evaluate the expression
$$\frac{1^2}{1^2-10+50}+\frac{2^2}{2^2-20+50}+\cdots+\frac{80^2}{80^2-80+50}$$
What should I do after that ?
$\sum\frac{n^2}{n^2-10n+50}$
I'm not seeing anything to find some way to cancel out the terms or something like that !
You can do this by approximating the sum by an integral: $$ \int_0^n \frac{x^2}{x^2-10x+50}\,dx $$ Mathematica says the integral is $$ n-5\ln 50+5\ln(50-10n+n^2) $$ although I can't tell you why. (see first comment below, thanks @Tim Ratigan) For $n=80$, this evaluates to $80+\ln 18424351793=103.637$. This is only an approximation that is useful for large $n$. For $n=80$, the sum itself is $104.203$.
EDIT
The general formula you posted is $$ \frac{n^2}{n^2-10n+50} $$ When $n=80$, substituting $n$ for $80$ gives $$\frac{80^2}{80^2-80\cdot 10+80}=\frac{80^2}{80^2-800+50}$$ For $n=8$, this does give $-80$ in the denominator, but then, you won't have an $80^2$ in the numerator.