Question

The sum $ 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + ... $ upto $ \infty $ terms, is equal to

• A. \( 6e \)
• B. \( 4e \)
• C. \( 3e \)
• D. \( 2e \)

Hint:

Identify the general term of the series. In this case, the numerator is the sum of the first \( n \) odd numbers, which is \( n^2 \). Simplify the expression \( \frac{n^2}{n!} \) by writing \( n^2 = n(n-1) + n \) and splitting the sum into simpler series that can be related to the Taylor series expansion of \( e^x \) at \( x = 1 \), which is \( e = \sum_{n=0}^{\infty} \frac{1}{n!} \).

Answer 1

The Correct Option is D

The \( n^{th} \) term of the series (for \( n \ge 1 \)) is given by: \[ T_n = \frac{1 + 3 + 5 + \dots + (2n - 1)}{n!} \] The sum of the first \( n \) odd numbers is \( n^2 \). So, the numerator is \( n^2 \). \[ T_n = \frac{n^2}{n!} \] The given sum \( S \) can be written as: \[ S = 1 + \sum_{n=2}^{\infty} \frac{n^2}{n!} = \sum_{n=1}^{\infty} \frac{n^2}{n!} \] We can write \( n^2 = n(n - 1) + n \). \[ S = \sum_{n=1}^{\infty} \frac{n(n - 1) + n}{n!} = \sum_{n=1}^{\infty} \frac{n(n - 1)}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!} \] For the first sum, the terms for \( n = 1 \) are zero. So, we start from \( n = 2 \): \[ \sum_{n=2}^{\infty} \frac{n(n - 1)}{n!} = \sum_{n=2}^{\infty} \frac{n(n - 1)}{n(n - 1)(n - 2)!} = \sum_{n=2}^{\infty} \frac{1}{(n - 2)!} \] Let \( m = n - 2 \). When \( n = 2 \), \( m = 0 \). When \( n \rightarrow \infty \), \( m \rightarrow \infty \). \[ \sum_{m=0}^{\infty} \frac{1}{m!} = e \] For the second sum: \[ \sum_{n=1}^{\infty} \frac{n}{n!} = \sum_{n=1}^{\infty} \frac{n}{n(n - 1)!} = \sum_{n=1}^{\infty} \frac{1}{(n - 1)!} \] Let \( k = n - 1 \). When \( n = 1 \), \( k = 0 \). When \( n \rightarrow \infty \), \( k \rightarrow \infty \). \[ \sum_{k=0}^{\infty} \frac{1}{k!} = e \] Therefore, the sum \( S \) is: \[ S = e + e = 2e \]