Question 25

Mathematics Binomial Expansions (Theorem) Medium

The value of the series \[ \frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \cdots \] is:

(A) \(2e^{-2}\)
(B) \(e^{-2}\)
(C) \(e^{-1}\)
(D) \(2e^{-1}\)
✓ Correct Answer: Option C Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

The given series is: \[ \frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \cdots \] Rewriting the sum: \[ \sum_{n=1}^{\infty} \frac{2n + 1 - 1}{(2n! + 1)} \] \[ \sum_{n=1}^{\infty} \frac{2n + 1}{2n! + 1} \cdot \frac{1}{2n! + 1} \] The series can be simplified as: \[ \sum_{n=1}^{\infty} \frac{1}{2n! + 1} - \frac{1}{2n! + 1} \] This simplifies to the following: \[ 1 + 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} - \frac{1}{6!} + \cdots = e^{-1} \] Thus, the value of the series is: \[ {e^{-1}} \]

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