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Question-129102

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Question Number 129102 by AbdulAzizabir last updated on 12/Jan/21
Answered by Olaf last updated on 12/Jan/21
Σ_(k=1) ^n k^2 = {1, 5, 14, 30, 55,...} = ((n(n+1)(2n+1))/6) S = (2/3)((1/4)−(1/(20))+(1/(56))−(1/(120))+(1/(220))−...) S = (1/6)((1/1)−(1/5)+(1/(14))−(1/(30))+(1/(55))−...) S = (1/6)Σ_(n=1) ^∞ (((−1)^(n+1) )/((n(n+1)(2n+1))/6)) S = Σ_(n=1) ^∞ (((−1)^(n+1) )/(n(n+1)(2n+1))) S = Σ_(n=1) ^∞ (−1)^(n+1) [(1/n)+(1/(n+1))−(4/(2n+1))] S = −Σ_(n=1) ^∞ (((−1)^n )/n)+Σ_(n=1) ^∞ (((−1)^(n+1) )/(n+1))+4Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) S = −Σ_(n=1) ^∞ (((−1)^n )/n)+Σ_(n=1) ^∞ (((−1)^n )/n)+1+4Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) S = 4Σ_(n=1) ^∞ (((−1)^n )/(2n+1))+1 (1/(1+x^2 )) = Σ_(n=0) ^∞ (−1)^n x^(2n) arctanx = Σ_(n=0) ^∞ (((−1)^n )/(2n+1))x^(2n+1) arctan(1) = Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) = (π/4) and Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) = (π/4)−1 ⇒ S = 4((π/4)−1)+1 = π−3
∑nk=1k2={1,5,14,30,55,…}=n(n+1)(2n+1)6S=23(14−120+156−1120+1220−…)S=16(11−15+114−130+155−…)S=16∑∞n=1(−1)n+1n(n+1)(2n+1)6S=∑∞n=1(−1)n+1n(n+1)(2n+1)S=∑∞n=1(−1)n+1[1n+1n+1−42n+1]S=−∑∞n=1(−1)nn+∑∞n=1(−1)n+1n+1+4∑∞n=1(−1)n2n+1S=−∑∞n=1(−1)nn+∑∞n=1(−1)nn+1+4∑∞n=1(−1)n2n+1S=4∑∞n=1(−1)n2n+1+111+x2=∑∞n=0(−1)nx2narctanx=∑∞n=0(−1)n2n+1x2n+1arctan(1)=∑∞n=0(−1)n2n+1=π4and∑∞n=1(−1)n2n+1=π4−1⇒S=4(π4−1)+1=π−3

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