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Question Number 129102 by AbdulAzizabir last updated on 12/Jan/21

	Answered by Olaf last updated on 12/Jan/21
	$Σ_(k=1) ^n k^2 = {1, 5, 14, 30, 55,...} = ((n(n+1)(2n+1))/6) S = (2/3)((1/4)−(1/(20))+(1/(56))−(1/(120))+(1/(220))−...) S = (1/6)((1/1)−(1/5)+(1/(14))−(1/(30))+(1/(55))−...) S = (1/6)Σ_(n=1) ^∞ (((−1)^(n+1) )/((n(n+1)(2n+1))/6)) S = Σ_(n=1) ^∞ (((−1)^(n+1) )/(n(n+1)(2n+1))) S = Σ_(n=1) ^∞ (−1)^(n+1) [(1/n)+(1/(n+1))−(4/(2n+1))] S = −Σ_(n=1) ^∞ (((−1)^n )/n)+Σ_(n=1) ^∞ (((−1)^(n+1) )/(n+1))+4Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) S = −Σ_(n=1) ^∞ (((−1)^n )/n)+Σ_(n=1) ^∞ (((−1)^n )/n)+1+4Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) S = 4Σ_(n=1) ^∞ (((−1)^n )/(2n+1))+1 (1/(1+x^2 )) = Σ_(n=0) ^∞ (−1)^n x^(2n) arctanx = Σ_(n=0) ^∞ (((−1)^n )/(2n+1))x^(2n+1) arctan(1) = Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) = (π/4) and Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) = (π/4)−1 ⇒ S = 4((π/4)−1)+1 = π−3$
	$\underset{k = 1}{\sum^{n}} k^{2} = {1, 5, 14, 30, 55, . . .} = \frac{n (n + 1) (2 n + 1)}{6}$ $S = \frac{2}{3} (\frac{1}{4} - \frac{1}{20} + \frac{1}{56} - \frac{1}{120} + \frac{1}{220} - . . .)$ $S = \frac{1}{6} (\frac{1}{1} - \frac{1}{5} + \frac{1}{14} - \frac{1}{30} + \frac{1}{55} - . . .)$ $S = \frac{1}{6} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{\frac{n (n + 1) (2 n + 1)}{6}}$ $S = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{n (n + 1) (2 n + 1)}$ $S = \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} [\frac{1}{n} + \frac{1}{n + 1} - \frac{4}{2 n + 1}]$ $S = - \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n} + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{n + 1} + 4 \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1}$ $S = - \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n} + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n} + 1 + 4 \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1}$ $S = 4 \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1} + 1$ $\frac{1}{1 + x^{2}} = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} x^{2 n}$ $\arctan x = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1} x^{2 n + 1}$ $\arctan (1) = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1} = \frac{π}{4}$ $and \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1} = \frac{π}{4} - 1$ $\Rightarrow S = 4 (\frac{π}{4} - 1) + 1 = π - 3$
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