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Question Number 125864 by MathSh last updated on 14/Dec/20

(1/(1+1^2 +1^4 ))+(2/(1+2^2 +2^4 ))+(3/(1+3^2 +3^4 ))+...+((50)/(1+50^2 +50^4 ))=?

$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{4}} }+\frac{\mathrm{2}}{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{4}} }+\frac{\mathrm{3}}{\mathrm{1}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{4}} }+...+\frac{\mathrm{50}}{\mathrm{1}+\mathrm{50}^{\mathrm{2}} +\mathrm{50}^{\mathrm{4}} }=? \\ $$

Answered by Dwaipayan Shikari last updated on 14/Dec/20

Σ_(n=1) ^n (n/(n^4 +n^2 +1))=Σ_(n=1) ^n (n/((n^2 +1−n)(n^2 +n+1)))=(1/2)Σ^n (1/(n^2 −n+1))−(1/(n^2 +n+1)) =(1/2)((1/1)−(1/3)+(1/3)−(1/7)+(1/7)−...−(1/(n^2 +n+1))) =(1/2)(1−(1/(n^2 +n+1)))=((n^2 +n)/(2(n^2 +n+1))) n=50 then Sum is (1/2)(1−(1/(2551)))=((1275)/(2551))

$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}}{{n}^{\mathrm{4}} +{n}^{\mathrm{2}} +\mathrm{1}}=\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}}{\left({n}^{\mathrm{2}} +\mathrm{1}−{n}\right)\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\overset{{n}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} −{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{7}}−...−\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right)=\frac{{n}^{\mathrm{2}} +{n}}{\mathrm{2}\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)} \\ $$$${n}=\mathrm{50}\:\:\:\:\:\:{then}\:{Sum}\:{is}\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2551}}\right)=\frac{\mathrm{1275}}{\mathrm{2551}} \\ $$

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