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

Σ_(k=1) ^n ((4k)/(4k^4 +1)) = ?

∑nk=14k4k4+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))

∑nk=1(2k2+2k+1)−(2k2−2k+1)(2k2+2k+1)(2k2−2k+1)=∑nk=1(12k2−2k+1−12k2+2k+1)=∑nk=1(12k2−2k+1−12(k+1)2−2(k+1)+1)=1−12n2+2n+1=2n2+2n2n2+2n+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 ))

∑nk=14k4k4+1+4k2−4k2=∑nk=14k(2k2+1−2k)(2k2+1+2k)=∑nk=112k2−2k+1−12k2+2k+1=(1−15+15−113+...−12n2+2n+1)=2n(n+1)n2+(n+1)2

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