A) \[\frac{{{\pi }^{2}}}{24}\]
B) \[\frac{{{\pi }^{2}}}{3}\]
C) \[\frac{{{\pi }^{2}}}{6}\]
D) None of these
Correct Answer: C
Solution :
We have, \[\sum\limits_{r=1}^{\infty }{\frac{1}{{{(2r-1)}^{2}}}}=\frac{{{\pi }^{2}}}{8}\] \[\Rightarrow \] \[\frac{1}{{{1}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{5}^{2}}}+...\infty =\frac{{{\pi }^{2}}}{8}\] ?(i) Let \[x=\sum\limits_{r=1}^{\infty }{\frac{1}{{{r}^{2}}}=\frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+...\infty }\] \[\Rightarrow \] \[x=\left( \frac{1}{{{1}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{5}^{2}}}+...\infty \right)\] \[+\left( \frac{1}{{{2}^{2}}}+\frac{1}{{{4}^{2}}}+\frac{1}{{{6}^{2}}}+...\infty \right)\] \[\Rightarrow \] \[x=\frac{{{\pi }^{2}}}{8}+\frac{1}{4}\left[ \frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+...\infty \right]\] \[\Rightarrow \] \[x=\frac{{{\pi }^{2}}}{8}+\frac{1}{4}.x\] \[\Rightarrow \] \[x-\frac{x}{4}=\frac{{{\pi }^{2}}}{8}\] \[\Rightarrow \] \[\frac{3x}{4}=\frac{{{\pi }^{2}}}{8}\] \[\Rightarrow \] \[x=\frac{{{\pi }^{2}}}{6}\]
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