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If $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+$......upto $\infty$ = $\frac{\phi^2}{g}$, then $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+$.....upto $\infty$ will be

$(a)\;\frac{\phi^2}{9}\qquad(b)\;\frac{\phi^2}{6}\qquad(c)\;\frac{\phi^2}{12}\qquad(d)\;\frac{\phi^2}{18}$

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Answer : (c) $\frac{\phi^2}{12}$
Explanation : $\;\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\;...\;upto\;\infty$
$=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....\;-\;\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...$
$=\frac{\phi^2}{g}-\frac{1}{2^2}\;[1+\frac{1}{2^2}+\frac{1}{3^2}+...]$
$=\frac{\phi^2}{g}-\frac{1}{2^2}\;(\frac{\phi^2}{9})$
$=\frac{\phi^2}{12}.$
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