Mathematics · Sequence and Series

JEE Main 2025 — 8 April, Evening Shift — Question 31

If 114+124+134+…∞=π490\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\ldots \infty=\frac{\pi^{4}}{90}, 114+134+154+…∞=α\frac{1}{1^{4}}+\frac{1}{3^{4}}+\frac{1}{5^{4}}+\ldots \infty=\alpha 124+144+164+…\frac{1}{2^{4}}+\frac{1}{4^{4}}+\frac{1}{6^{4}}+\ldots

∞=β \infty=\beta, Then αβ\frac{\alpha}{\beta} is equal to

  1. Option A:

    14

  2. Option B:

    15

    Correct
  3. Option C:

    18

  4. Option D:

    23

Answer: B

Step-by-step solution

α=114+134+154+…\alpha=\frac{1}{1^{4}}+\frac{1}{3^{4}}+\frac{1}{5^{4}}+\ldots

β=124+144+164+…=124[114+124+134+144+…]\begin{aligned} & \beta=\frac{1}{2^{4}}+\frac{1}{4^{4}}+\frac{1}{6^{4}}+\ldots \\& =\frac{1}{2^{4}}\left[\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\frac{1}{4^{4}}+\ldots\right] \end{aligned}

⇒16β=[114+134+154+…]+[124+144+164+…]\Rightarrow 16 \beta=\left[\frac{1}{1^{4}}+\frac{1}{3^{4}}+\frac{1}{5^{4}}+\ldots\right]+\left[\frac{1}{2^{4}}+\frac{1}{4^{4}}+\frac{1}{6^{4}}+\ldots\right] =α+β=\alpha+\beta

⇒15β=α⇒αβ=15 \Rightarrow 15 \beta=\alpha \Rightarrow \frac{\alpha}{\beta}=15

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Mathematics
Chapter
Sequence and Series
Topic
Introduction to Sequence and Series