If , , , then is equal to [2025]

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Solution

Q.
If $\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + . . . \infty = \frac{π^{4}}{90}$ , $\frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + . . . \infty = α$ , $\frac{1}{2^{4}} + \frac{1}{4^{4}} + \frac{1}{6^{4}} + . . . \infty = β$ , then $\frac{α}{β}$ is equal to [2025]

1 14

2 23

3 15

4 18

Ans.
(3)

We have, $β = \frac{1}{2^{4}} + \frac{1}{4^{4}} + \frac{1}{6^{4}} + . . .$

$= \frac{1}{2^{4}} (1 + \frac{1}{2^{4}} + \frac{1}{3^{4}} + . . .) = \frac{1}{16} (\frac{π^{4}}{90})$ [Given]

$∴ β = \frac{π^{4}}{1440}$ ... (i)

Now, $\frac{π^{4}}{90} = \frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + . . .$

$= (\frac{1}{1^{4}} + \frac{1}{3^{4}} + \frac{1}{5^{4}} + . . . . . .) + (\frac{1}{2^{4}} + \frac{1}{4^{4}} + . . . . . .)$

$= α + β$

$\Rightarrow α = \frac{π^{4}}{90} - β = \frac{π^{4}}{90} - \frac{π^{4}}{1440}$ [Using (i)]

$= \frac{16 π^{4} - π^{4}}{1440} = \frac{15 π^{4}}{1440}$

$∴ \frac{α}{β} = \frac{15}{1} = 15$ .

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