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Question Number 204642 by BaliramKumar last updated on 24/Feb/24

If \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \frac{1}{5^{2}} + \dots \dots \dots \dots . = \frac{π^{2}}{6}

then \frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \dots \dots \dots \dots . = ?

Answered by Rasheed.Sindhi last updated on 24/Feb/24

\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \frac{1}{5^{2}} + \dots \dots \dots \dots . = \frac{π^{2}}{6}

(\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \cdot \cdot \cdot) + (\frac{1}{2^{2}} + \frac{1}{4^{2}} + \frac{1}{6^{2}} + \cdot \cdot \cdot) = \frac{π^{2}}{6}

(\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \cdot \cdot \cdot) + \frac{1}{2^{2}} (\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \cdot \cdot \cdot) = \frac{π^{2}}{6}

(\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \cdot \cdot \cdot) + \frac{1}{4} (\frac{π^{2}}{6}) = \frac{π^{2}}{6}

(\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \cdot \cdot \cdot) = \frac{π^{2}}{6} - \frac{π^{2}}{24} = \frac{4 π^{2} - π^{2}}{24} = \frac{π^{2}}{8}

Commented by BaliramKumar last updated on 24/Feb/24

Thanks