If 11^4+12^4+13^4+ = ^490, 11^4+13^4+15^4+ = , 12^4+14^4+16^4+ = , then ( )/( ) is equal t…

Question

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 \infty=\alpha $,
 $\frac{1}{{2}^{4}}+\frac{1}{{4}^{4}}+\frac{1}{{6}^{4}}+\ldots \infty=\beta $,
 then $\frac{\alpha }{\beta }$ is equal to :

Accepted Answer

$$\begin{aligned}
& \text { If } \frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots . \infty=\frac{\pi^4}{90} \quad\text{....... (i)}\
& \beta=\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\ldots, \
& =\frac{1}{16}\left[\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots . .\right],
\end{aligned}$$

$=\frac{1}{16} \times \frac{\pi^4}{90}$ using (ii) ............ (ii)

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

Solutions · 1