જો sumlimits_{{text{r}}, = ,{text{1}}}^i | vedclass

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Question

અહી, $\,\,\frac{1}{{{1^2}}}\,\, + \,\,\frac{1}{{\,{3^2}}}\,\, + \,\,\frac{1}{{{5^2}}}\,\, + \,.....\,\,\infty \,\, = \,\frac{{{\pi ^2}}}{8} $

Vedclass · Accepted Answer

$\frac{{{\pi ^2}}}{6}$

Vedclass · Answer

અહી, $\,\,\frac{1}{{{1^2}}}\,\, + \,\,\frac{1}{{\,{3^2}}}\,\, + \,\,\frac{1}{{{5^2}}}\,\, + \,.....\,\,\infty \,\, = \,\frac{{{\pi ^2}}}{8} $

$x\,\,=\,\sum\limits_{{	ext{r}}\, = \,{	ext{1}}}^\infty  {\frac{1}{{{r^2}}}} $ લેતાં,

$x=\frac{1}{{{1^2}}}\,\, + \,\,\frac{1}{{{2^2}}}\,\, + \,\,\frac{1}{{{3^2}}}\,\, + \,\,.....\,\,\infty $ 

$x = \,\,\left( {\frac{1}{{{1^2}}}\,\, + \,\,\frac{1}{{{3^2}}}\,\, + \,\,\frac{1}{{\,{5^2}}}\, + \,.....\,\infty } ight)\, + \,\left( {\frac{1}{{{2^2}}}\,\, + \,\,\frac{1}{{{4^2}}}\,\, + \,\,\frac{1}{{{6^2}}}\,\, + \,\,.....\,\,\infty } ight)$

$x = \,\,\frac{{{\pi ^2}}}{8}\, + \,\,\frac{1}{4}\,\left( {\frac{1}{{{1^2}}}\,\, + \,\,\frac{1}{{{2^2}}}\,\, + \,\,\frac{1}{{{3^2}}}\,\, + ....\,\,\infty } ight)\,\,$

$x= \,\,\frac{{{\pi ^2}}}{8}\,\, + \,\,\frac{1}{4}\,x$

$x= \,\,\frac{{{\pi ^2}}}{6}$

જો $\sum\limits_{{\text{r}}\, = \,{\text{1}}}^\infty {\frac{1}{{{{(2r\, - \,1)}^2}}}\,\, = \,\,\frac{{{\pi ^2}}}{8}} $ હોય, તો $\,\sum\limits_{{\text{r}}\, = \,{\text{1}}}^\infty {\frac{1}{{{r^2}}}\,\, = \,\,.........} $

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