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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}}}\,\, = \,\,.........} $
$\frac{{{\pi ^2}}}{{24}}$
$\frac{{{\pi ^2}}}{3}$
$\frac{{{\pi ^2}}}{6}$
આપેલ પૈકી એકપણ નહિ.
Solution
અહી, $\,\,\frac{1}{{{1^2}}}\,\, + \,\,\frac{1}{{\,{3^2}}}\,\, + \,\,\frac{1}{{{5^2}}}\,\, + \,…..\,\,\infty \,\, = \,\frac{{{\pi ^2}}}{8} $
$x\,\,=\,\sum\limits_{{\text{r}}\, = \,{\text{1}}}^\infty {\frac{1}{{{r^2}}}} $ લેતાં,
$x=\frac{1}{{{1^2}}}\,\, + \,\,\frac{1}{{{2^2}}}\,\, + \,\,\frac{1}{{{3^2}}}\,\, + \,\,…..\,\,\infty $
$x = \,\,\left( {\frac{1}{{{1^2}}}\,\, + \,\,\frac{1}{{{3^2}}}\,\, + \,\,\frac{1}{{\,{5^2}}}\, + \,…..\,\infty } \right)\, + \,\left( {\frac{1}{{{2^2}}}\,\, + \,\,\frac{1}{{{4^2}}}\,\, + \,\,\frac{1}{{{6^2}}}\,\, + \,\,…..\,\,\infty } \right)$
$x = \,\,\frac{{{\pi ^2}}}{8}\, + \,\,\frac{1}{4}\,\left( {\frac{1}{{{1^2}}}\,\, + \,\,\frac{1}{{{2^2}}}\,\, + \,\,\frac{1}{{{3^2}}}\,\, + ….\,\,\infty } \right)\,\,$
$x= \,\,\frac{{{\pi ^2}}}{8}\,\, + \,\,\frac{1}{4}\,x$
$x= \,\,\frac{{{\pi ^2}}}{6}$
