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Question : 16 of 90

Marks: +1, -0

Given that

1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots \ldots=\frac{\pi^{2}}{6}

1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+\ldots \ldots

. is

$\frac{\pi^{2}}{6}-1$
$\frac{\pi}{6}$
$\frac{\pi^{2}}{12}$
$\frac{\pi^{2}}{8}$

Solution:

Given,

1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots \ldots=\frac{\pi^{2}}{6} \longrightarrow(1)

Multiplying all the terms by

\frac{1}{4}

we get,

\begin{aligned} & \frac{1}{4}+\frac{1}{2^{2} \cdot 4}+\frac{1}{3^{2} \cdot 4}+\frac{1}{4^{2} \cdot 4}+\ldots \ldots=\frac{\pi^{2}}{6 \cdot 4}=\frac{\pi^{2}}{24} \\ & \text { or, } \frac{1}{2^{2}}+\frac{1}{2^{2} \cdot 2^{2}}+\frac{1}{3^{2} \cdot 2^{2}}+\frac{1}{4^{2} \cdot 2^{2}}+\ldots \ldots=\frac{\pi^{2}}{6 \cdot 4}=\frac{\pi^{2}}{24} \\ & \text { or, } \frac{1}{2^{2}}+\frac{1}{4^{2}}+\frac{1}{6^{2}}+\frac{1}{8^{2}}+\ldots \ldots=\frac{\pi^{2}}{6 \cdot 4}=\frac{\pi^{2}}{24} \longrightarrow \text { (2) } \end{aligned}

Subtracting (2) from (1),

1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+\ldots \ldots=\frac{\pi^{2}}{6}-\frac{\pi^{2}}{24}=\frac{3 \pi^{2}}{24}=\frac{\pi^{2}}{8}

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