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Algebra Level 4

$\large 1 + \frac{1}{2^2} + \frac{1}{3^2} + ... = \frac{\pi^2}{6}$

Given the above, find $n$ that satisfies the equation below.

$\large 1 + \frac{1}{3^2} + \frac{1}{5^2} + ... = \frac{\pi^2}{n}$

The answer is 8.

2 solutions

Chew-Seong Cheong
Nov 14, 2016

$\begin{aligned} S & = 1 + \frac 1{3^2} + \frac 1{5^2} + \frac 1{7^2} + ... \\ & = 1 + {\color{#D61F06}\frac 1{2^2}} + \frac 1{3^2} + {\color{#D61F06}\frac 1{4^2}} + \frac 1{5^2} + {\color{#D61F06}\frac 1{6^2}} + \frac 1{7^2} + ... - \left( {\color{#D61F06}\frac 1{2^2}} + {\color{#D61F06}\frac 1{4^2}} + {\color{#D61F06}\frac 1{6^2}} + ...\right) \\ & = 1 + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} + ... - \frac 1{2^2} \left(1 + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} + ... \right) \\ & = \left(1 - \frac 14 \right) \frac {\pi^2}6 \\ & = \frac 34 \cdot \frac {\pi^2}6 \\ & = \frac {\pi^2}8 \end{aligned}$

$\implies n = \boxed{8}$

@Priyanshu Mishra , please don't use "huge" in your question. It makes it look childish. Just use the standard three dots " $...$ " and no more no less.

Chew-Seong Cheong - 4 years, 7 months ago

Okay.

I wanted the solution. Yours one is very good.

Priyanshu Mishra - 4 years, 7 months ago

Viki Zeta
Nov 14, 2016

$1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \ldots = \sum \ \dfrac{1}{n^2} = \dfrac{\pi^2}{6} \\ x = \sum \ \dfrac{1}{n^2} = \dfrac{\pi^2}{6} \\ \dfrac{x}{4} = \dfrac{1}{4} \sum \ \dfrac{1}{n^2} = \dfrac{\pi^2}{24}\\ \ \ \ \ \ x = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \ldots\\ - \dfrac{x}{4} = \dfrac{1}{2^2} + \dfrac{1}{4^2} + \dfrac{1}{6^2} + \ldots \\ \dfrac{\pi^2}{6} - \dfrac{\pi^2}{24} = 1 + \dfrac{1}{3^2} + \dfrac{1}{5^2} + \ldots \\ \boxed{1 + \dfrac{1}{3^2} + \dfrac{1}{5^2} + \ldots = \dfrac{\pi^2}{8} \\ \therefore n = 8}$

Have seen before

The answer is 8.

2 solutions

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