aligned & If (1)/(1^4)+(1)/(2^4)+(1)/(3^4)+ . . ∞=(π^4)/(90), & (1)/(1^4)+(1)/(3^4)+(1)/(5^4)+ . . ∞=α, & (1)/(2^4)+(1)/(4^4)+(1)/(6^4)+ . ∞=β, aligned then (α)/(β) is equal to

Correct answer: C. 15 If (1)/(1^4)+(1)/(2^4)+(1)/(3^4)+ . . ∞=(π^4)/(90)....(i) aligned & β=(1)/(2^4)+(1)/(4^4)+(1)/(6^4)+ , & =(1)/(16)[(1)/(1^4)+(1)/(2^4)+(1)/(3^4)+ . .], aligned =(1)/(16) × (π^4)/(90) using (ii) ________...(ii) aligned & α=(1)/(1^4)+(1)/(3^4)+(1)/(5^4)+ . . ∞ & ((1)/(1^4)+(1)/(2^4)+(1)/(3^4)+(1)/(4^4)+(1)/(5^4)+ . .) & -((1)/(2^4)+(1)/(4^4)+(1)/(6^4)+ . .) aligned α=(π^4)/(90)-(1)/(16) × (π^4)/(90) [ using (i) and (ii)] aligned & α=(16-1)/(16 × 90) × π^4=(15)/(16 × 90) π^4=(π^4)/(96) & (α)/(β)=((π^4)/(96))/((π^4)/(16 × 90))=(16 × 90)/(96)=15 aligned

JEE Main 20258 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Sequences and Series

aligned & If (1)/(1^4)+(1)/(2^4)+(1)/(3^4)+ . . ∞=(π^4)/(90), &...

\begin{aligned}& \text{ 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,\end{aligned}

then

\frac{\alpha}{\beta}

is equal to

Options

Correct answer

C. 15

Step-by-step solution

\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots . . \infty=\frac{\pi^4}{90}....(i)

\begin{aligned} & \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)\end{aligned}

\alpha=\frac{\pi^4}{90}-\frac{1}{16} \times \frac{\pi^4}{90} \quad[

using (i) and (ii)]

\begin{aligned} & \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}

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