JEE Mathematics 2025 Question with Solution

Given:

• $S = \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \dots = \frac{\pi^4}{90}$
• $\alpha = \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \dots$
• $\beta = \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \dots$

Find: $\frac{\alpha}{\beta}$

The complete series is split into odd and even terms, so

$$S = \alpha + \beta$$

Hence,

$$\alpha + \beta = \frac{\pi^4}{90}$$

For the even-term series,

$$\beta = \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \dots$$

Factor out $\frac{1}{2^4} = \frac{1}{16}$:

$$\beta = \frac{1}{16}\left(1 + \frac{1}{2^4} + \frac{1}{3^4} + \dots\right)$$

So,

$$\beta = \frac{1}{16} S = \frac{1}{16} \cdot \frac{\pi^4}{90} = \frac{\pi^4}{1440}$$

Now,

$$\alpha = \frac{\pi^4}{90} - \frac{\pi^4}{1440}$$

Using a common denominator,

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

Therefore,

$$\frac{\alpha}{\beta} = \frac{\frac{15\pi^4}{1440}}{\frac{\pi^4}{1440}} = 15$$

So, the correct option is B.

Given: The full reciprocal fourth-power series is $\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$.

Find: The ratio $\frac{\alpha}{\beta}$ where $\alpha$ is the sum over odd terms and $\beta$ is the sum over even terms.

Write the total series as

$$S = \alpha + \beta$$

with

$$\alpha = \sum_{n\,\text{odd}} \frac{1}{n^4}, \qquad \beta = \sum_{n\,\text{even}} \frac{1}{n^4}$$

Now express the even series by putting $n = 2k$:

$$\beta = \sum_{k=1}^{\infty} \frac{1}{(2k)^4}$$ $$\beta = \sum_{k=1}^{\infty} \frac{1}{16k^4} = \frac{1}{16} \sum_{k=1}^{\infty} \frac{1}{k^4}$$

Therefore,

$$\beta = \frac{1}{16} \cdot \frac{\pi^4}{90} = \frac{\pi^4}{1440}$$

Then,

$$\alpha = S - \beta = \frac{\pi^4}{90} - \frac{\pi^4}{1440}$$ $$\alpha = \frac{16\pi^4 - \pi^4}{1440} = \frac{15\pi^4}{1440} = \frac{\pi^4}{96}$$

Hence,

$$\frac{\alpha}{\beta} = \frac{\pi^4/96}{\pi^4/1440} = \frac{1440}{96} = 15$$

Therefore, the ratio is $15$, so the correct option is B.