Evaluate the expression1212−10+50+2222−20+50+⋯+802802−80+50What should I do after that ?∑n2n2−10n+50I'm not seeing anything to find some way to cancel out the terms or something like that !

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Evaluate the expression1212−10+50+2222−20+50+⋯+802802−80+50What should I do after that ?∑n2n2−10n+50I&#039;m not seeing anything to find some way to cancel out the terms or something like that !

Denise Daniel · Accepted Answer

You can do this by approximating the sum by an integral:∫0nx2x2−10x+50dxMathematica says the integral isn−5ln⁡50+5ln⁡(50−10n+n2)For n=80, this evaluates to 80+ln⁡18424351793=103.637This is only an approximation that is useful for large n. For n=80, the sum itself is 104.203.EDITThe general formula you posted isn2n2−10n+50When n=80, substituting n for 80 gives802802−80⋅10+80=802802−800+50For n=8, this does give −80 in the denominator, but then, you won&#039;t have an 802 in the numerator.

Evaluate the expression \(\displaystyle{\frac{{{1}^{{2}}}}{{{1}^{{2}}-{10}+{50}}}}+{\frac{{{2}^{{2}}}}{{{2}^{{2}}-{20}+{50}}}}+\cdots+{\frac{{{80}^{{2}}}}{{{80}^{{2}}-{80}+{50}}}}\) What should I do after

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