Question Number 175217 by mnjuly1970 last updated on 23/Aug/22
Question Number 109639 by Ar Brandon last updated on 24/Aug/20 Answered by Dwaipayan Shikari last updated on 25/Aug/20 $${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }=\left(−\mathrm{1}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}}…
Question Number 44085 by rk last updated on 21/Sep/18
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Question Number 175115 by mnjuly1970 last updated on 19/Aug/22 Answered by aleks041103 last updated on 19/Aug/22
Question Number 109462 by 150505R last updated on 23/Aug/20 Answered by mathmax by abdo last updated on 24/Aug/20 $$\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left[\mathrm{k}^{\mathrm{2}} \mathrm{x}\:+\mathrm{k}^{\mathrm{2}} \right]\:\Rightarrow\:\mathrm{A}_{\mathrm{n}}…
Question Number 174951 by infinityaction last updated on 14/Aug/22
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