If 1 1^ 4 +1 2^ 4 +1 3^ 4 + . . = ^ 4 90 , 1 1^ 4 +1 3^ 4 +1 5^ 4… - Vidyadip

MCQ

If $\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\ldots . . \infty=\frac{\pi^{4}}{90}$,
$\frac{1}{1^{4}}+\frac{1}{3^{4}}+\frac{1}{5^{4}}+\ldots \ldots \infty=\alpha$,
$\frac{1}{2^{4}}+\frac{1}{4^{4}}+\frac{1}{6^{4}}+\ldots . . \infty=\beta$,
then $\frac{\alpha}{\beta}$ is equal to

A
23
B
18
✓
15
D
14

✓

Answer

Correct option: C.

15

(C) 15
If $\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\ldots . . \infty=\frac{\pi^{4}}{90}$ $\ldots(i)$
$\beta=\frac{1}{2^{4}}+\frac{1}{4^{4}}+\frac{1}{6^{4}}+\ldots$,
$=\frac{1}{16}\left[\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\ldots ..\right]$,
$=\frac{1}{16} \times \frac{\pi^{4}}{90} \quad$ using (ii) $\ldots(ii)$
$\alpha=\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\ldots \ldots \infty$
$\left(\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\frac{1}{4^{4}}+\frac{1}{5^{4}}+\ldots ..\right)$
$-\left(\frac{1}{2^{4}}+\frac{1}{4^{4}}+\frac{1}{6^{4}}+\ldots ..\right)$
$\alpha=\frac{\pi^{4}}{90}-\frac{1}{16} \times \frac{\pi^{4}}{90} \quad[$ using (i) and (ii)]
$\alpha=\frac{16-1}{16 \times 90} \times \pi^{4}=\frac{15}{16 \times 90} \pi^{4}=\frac{\pi^{4}}{96}$
$\therefore \frac{\alpha}{\beta}=\frac{\frac{\pi^{4}}{96}}{\frac{\pi^{4}}{16 \times 90}}=\frac{16 \times 90}{96}=15$

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