If `S_(n) = (1^(2).2)/(1!) + (2^(2).3)/(2!) + (3^(2).4)/(3!) + ... + (n^(2)(n+1))/(n!)`, then `lim_(n to oo) sum S_(n)` is equal to

Options

A. 3e
B. 5e
C. 7e
D. 9e

Answer

To solve the problem, we need to evaluate the limit of the sum \( S_n \) as \( n \) approaches infinity, where

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

### Step 1: Rewrite the expression for \( S_n \)

We can express \( S_n \) in a more manageable form:

\[ S_n = \sum_{k=1}^{n} \frac{k^2 (k+1)}{k!} = \sum_{k=1}^{n} \left( \frac{k^3}{k!} + \frac{k^2}{k!} \right). \]

### Step 2: Separate the sums

Now, we can separate the sums:

\[ S_n = \sum_{k=1}^{n} \frac{k^3}{k!} + \sum_{k=1}^{n} \frac{k^2}{k!}. \]

### Step 3: Evaluate the first sum

To evaluate \( \sum_{k=1}^{\infty} \frac{k^3}{k!} \), we can use the known result that:

\[ \sum_{k=0}^{\infty} \frac{k^m}{k!} = e \quad \text{for } m = 0, 1, 2, \ldots \]

Specifically, we can derive that:

\[ \sum_{k=0}^{\infty} \frac{k^3}{k!} = e \cdot (1 + 1 + 1) = 3e. \]

### Step 4: Evaluate the second sum

For the second sum, we similarly know:

\[ \sum_{k=0}^{\infty} \frac{k^2}{k!} = e \cdot (1 + 1) = 2e. \]

### Step 5: Combine the results

Now we can combine the results of both sums:

\[ \sum_{k=1}^{\infty} \frac{k^3}{k!} + \sum_{k=1}^{\infty} \frac{k^2}{k!} = 3e + 2e = 5e. \]

### Step 6: Take the limit

Thus, we find that:

\[ \lim_{n \to \infty} S_n = 5e. \]

### Final Answer

The limit is:

\[ \lim_{n \to \infty} \sum S_n = 5e. \]