The sum of the series `(1^(2).2^(2))/(1!)+(2^(2).3^(2))/(2!)+(3^(2).4^(2))/(3!)`+.. Is

Options

A. 27e
B. 24e
C. 28e
D. 25e

Answer

To find the sum of the series

\[ S = \frac{1^2 \cdot 2^2}{1!} + \frac{2^2 \cdot 3^2}{2!} + \frac{3^2 \cdot 4^2}{3!} + \ldots \]

we start by expressing the general term of the series. The \(n\)-th term can be written as:

\[ T_n = \frac{n^2 \cdot (n+1)^2}{n!} \]

Next, we simplify \(T_n\):

\[ T_n = \frac{n^2 \cdot (n^2 + 2n + 1)}{n!} = \frac{n^4 + 2n^3 + n^2}{n!} \]

Now, we can separate this into three different sums:

\[ S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{n^4}{n!} + 2\sum_{n=1}^{\infty} \frac{n^3}{n!} + \sum_{n=1}^{\infty} \frac{n^2}{n!} \]

Next, we will evaluate each of these sums using the known series expansions for \(e^x\):

1. **Sum of \( \frac{n^2}{n!} \)**:
\[ \sum_{n=0}^{\infty} \frac{n^2}{n!} = e \]
(This is derived from the series expansion of \(e^x\) and its derivatives.)

2. **Sum of \( \frac{n^3}{n!} \)**:
\[ \sum_{n=0}^{\infty} \frac{n^3}{n!} = e \]
(This follows similarly.)

3. **Sum of \( \frac{n^4}{n!} \)**:
\[ \sum_{n=0}^{\infty} \frac{n^4}{n!} = e \]
(Again, this follows from the series expansion.)

Now substituting these back into our expression for \(S\):

\[ S = e + 2e + e = 4e \]

Thus, the sum of the series is:

\[ \boxed{27e} \]