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Question Number 122850 by liberty last updated on 20/Nov/20

Σ_(k=1) ^n ((4k)/(4k^4 +1)) = ?

$$\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{4}{k}}{\mathrm{4}{k}^{\mathrm{4}} +\mathrm{1}}\:=\:? \\ $$

Answered by bemath last updated on 20/Nov/20

Σ_(k=1) ^n (((2k^2 +2k+1)−(2k^2 −2k+1))/((2k^2 +2k+1)(2k^2 −2k+1))) = Σ_(k=1) ^n ((1/(2k^2 −2k+1)) − (1/(2k^2 +2k+1))) = Σ_(k=1) ^n ((1/(2k^2 −2k+1)) − (1/(2(k+1)^2 −2(k+1)+1)))= 1−(1/(2n^2 +2n+1)) = ((2n^2 +2n)/(2n^2 +2n+1))

$$\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\left(\mathrm{2}{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}\right)−\left(\mathrm{2}{k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{1}\right)}{\left(\mathrm{2}{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{1}\right)}\:= \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{1}}\:−\:\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}}\right)\:= \\ $$$$\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{1}}\:−\:\frac{\mathrm{1}}{\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({k}+\mathrm{1}\right)+\mathrm{1}}\right)= \\ $$$$\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}}\:=\:\frac{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{2}{n}}{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}} \\ $$

Answered by Dwaipayan Shikari last updated on 20/Nov/20

Σ_(k=1) ^n ((4k)/(4k^4 +1+4k^2 −4k^2 )) =Σ_(k=1) ^n ((4k)/((2k^2 +1−2k)(2k^2 +1+2k))) =Σ_(k=1) ^n (1/(2k^2 −2k+1))−(1/(2k^2 +2k+1)) =(1−(1/5)+(1/5)−(1/(13))+...−(1/(2n^2 +2n+1))) =((2n(n+1))/(n^2 +(n+1)^2 ))

$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{4}{k}}{\mathrm{4}{k}^{\mathrm{4}} +\mathrm{1}+\mathrm{4}{k}^{\mathrm{2}} −\mathrm{4}{k}^{\mathrm{2}} } \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{4}{k}}{\left(\mathrm{2}{k}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{k}\right)\left(\mathrm{2}{k}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{k}\right)} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{1}} \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{13}}+...−\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{2}{n}\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} +\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

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