Discussion

Explanation:

If $\frac{1}{1^2}$ + $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + ... upto ∞ = $\frac{{\mathrm{\pi }}^2}{6}$, then the value of 

Let X = $\frac{1}{1^2}$ + $\frac{1}{3^2}$ + $\frac{1}{5^2}$ + ... upto ∞

Adding and subtracting $\frac{1}{2^2}$ + $\frac{1}{4^2}$ + $\frac{1}{6^2}$ + ... upto ∞, we have

X = $\frac{1}{1^2}$ + $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + ... upto ∞ - ($\frac{1}{2^2}$ + $\frac{1}{4^2}$ + $\frac{1}{6^2}$ + ... upto ∞)

⇒ X = $\frac{{\mathrm{\pi }}^2}{6}$ - $\frac{1}{2^2}$($\frac{1}{1^2}$ + $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + ... upto ∞)

⇒ X = $\frac{{\mathrm{\pi }}^2}{6}$ - $\frac{1}{2^2}$($\frac{{\mathrm{\pi }}^2}{6}$)

⇒ X = $\frac{{\mathrm{\pi }}^2}{6}$ - $\frac{1}{4}$($\frac{{\mathrm{\pi }}^2}{6}$)

⇒ X = $\frac{{\mathrm{\pi }}^2}{8}$

Hence, option (a).

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