Question

The sum of the series \[ 1 + \frac{1^2 + 2^2}{2!} + \frac{1^2 + 2^2 + 3^2}{3!} + \frac{1^2 + 2^2 + 3^2 + 4^2}{4!} + \dots \] is:

• A. \( 3e \)
• B. \( \frac{17}{6}e \)
• C. \( \frac{13}{6}e \)
• D. \( 19e \)

Hint:

To evaluate such series, recognize the general form of the terms and use known summation formulas, such as the sum of squares, and standard series expansions for functions like \( e^x \).

Answer 1

The Correct Option is B

We are given the series: \[ S = 1 + \frac{1^2 + 2^2}{2!} + \frac{1^2 + 2^2 + 3^2}{3!} + \frac{1^2 + 2^2 + 3^2 + 4^2}{4!} + \dots \] We want to find the sum of this series. Let’s break it down and find a general form for the \( n \)-th term. Step 1: General term of the series. The general \( n \)-th term of the series is: \[ T_n = \frac{1^2 + 2^2 + \dots + n^2}{n!}. \] The sum of squares from \( 1^2 \) to \( n^2 \) is given by the formula: \[ \sum_{k=1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}. \] So, the general term becomes: \[ T_n = \frac{\frac{n(n + 1)(2n + 1)}{6}}{n!} = \frac{n(n + 1)(2n + 1)}{6n!}. \] Step 2: Summing the series. The sum of the series is: \[ S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{n(n + 1)(2n + 1)}{6n!}. \] Now, this series can be simplified and recognized as a standard series related to the exponential function \( e \). The final result gives us the sum as \( \frac{17}{6}e \). Step 3: Conclusion. Therefore, the sum of the series is \( \frac{17}{6}e \), and the correct answer is (b).