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Question-134814




Question Number 134814 by Ndala last updated on 07/Mar/21
Commented by Ndala last updated on 07/Mar/21
I need your help my friends!
$$\mathrm{I}\:\mathrm{need}\:\mathrm{your}\:\mathrm{help}\:\mathrm{my}\:\mathrm{friends}! \\ $$
Answered by Ar Brandon last updated on 07/Mar/21
Σ_(n=1) ^(50) (n/(1+n^2 +n^4 ))=Σ_(n=1) ^(50) (n/((n^2 −n+1)(n^2 +n+1)))  (n/((n^2 −n+1)(n^2 +n+1)))=((an+b)/(n^2 −n+1))+((cn+d)/(n^2 +n+1))  Partial fraction. Then answer can be found  through telescopy.
$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{50}} {\sum}}\frac{\mathrm{n}}{\mathrm{1}+\mathrm{n}^{\mathrm{2}} +\mathrm{n}^{\mathrm{4}} }=\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{50}} {\sum}}\frac{\mathrm{n}}{\left(\mathrm{n}^{\mathrm{2}} −\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{n}}{\left(\mathrm{n}^{\mathrm{2}} −\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{1}\right)}=\frac{\mathrm{an}+\mathrm{b}}{\mathrm{n}^{\mathrm{2}} −\mathrm{n}+\mathrm{1}}+\frac{\mathrm{cn}+\mathrm{d}}{\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{1}} \\ $$$$\mathrm{Partial}\:\mathrm{fraction}.\:\mathrm{Then}\:\mathrm{answer}\:\mathrm{can}\:\mathrm{be}\:\mathrm{found} \\ $$$$\mathrm{through}\:\mathrm{telescopy}. \\ $$
Commented by Ndala last updated on 07/Mar/21
Thanks sir!
$$\mathrm{Thanks}\:\mathrm{sir}! \\ $$

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