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