Sequences & Series
Sequence and Series
star_batch_jee_advanced_2025
Grade None
Question:
The value of $\frac{(1^4 + \frac{1}{4})(3^4 + \frac{1}{4})...(2n-1)^4 + \frac{1}{4})}{(2^4 + \frac{1}{4})(4^4 + \frac{1}{4})...(2n)^4 + \frac{1}{4})}$ is equal to:
$\frac{1}{4n^2 + 2n + 1}$
$\frac{1}{8n^2 + 4n + 1}$
$\frac{1}{4(2n^2 + n + 1)}$
$\frac{n}{8n^2 - 4n + 1}$
Step-by-Step Solution
Key Concept: Telescoping products arise when numerators and denominators of consecutive factors share common expressions.
Using the identity $k^4 + \frac{1}{4} = (k^2 - k + \frac{1}{2})(k^2 + k + \frac{1}{2})$, the product $\prod_{k=1}^{n} \frac{(2k-1)^2 - (2k-1) + \frac{1}{2}}{(2k)^2 - 2k + \frac{1}{2}} \cdot \frac{(2k-1)^2 + (2k-1) + \frac{1}{2}}{(2k)^2 + 2k + \frac{1}{2}}$ telescopes. The result simplifies to $\frac{1}{4n^2 + 2n + 1} \cdot \frac{1}{8n^2 + 4n + 1}$.
Correct Answer: 2