`1.3 + (2.4)/(1.2) + (3.5)/(1.2.3) + (4.6)/(1.2.3.4) +...`

Options

A. e
B. 2e
C. 3e
D. 4e

Answer

To solve the series given by

\[ S = 1.3 + \frac{2.4}{1.2} + \frac{3.5}{1.2.3} + \frac{4.6}{1.2.3.4} + \ldots \]

we will first express the general term of the series and then find its sum.

### Step 1: Identify the General Term

The series can be rewritten in a more general form. The \( n \)-th term can be expressed as:

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

where \( n! \) is the factorial of \( n \).

### Step 2: Simplify the General Term

We can simplify \( T_n \):

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

This can be further broken down into:

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

### Step 3: Rewrite the Series

Now we can express the series \( S \) as:

\[ S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \left( \frac{n}{(n-1)!} + \frac{2}{(n-1)!} \right) \]

### Step 4: Change the Index of Summation

To simplify the summation, we can change the index of summation. Let \( m = n - 1 \), then \( n = m + 1 \) and the series becomes:

\[ S = \sum_{m=0}^{\infty} \left( \frac{m+1}{m!} + \frac{2}{m!} \right) \]

This can be separated into two sums:

\[ S = \sum_{m=0}^{\infty} \frac{m+1}{m!} + 2 \sum_{m=0}^{\infty} \frac{1}{m!} \]

### Step 5: Evaluate Each Sum

1. The first sum can be evaluated as follows:

\[ \sum_{m=0}^{\infty} \frac{m+1}{m!} = \sum_{m=0}^{\infty} \frac{m}{m!} + \sum_{m=0}^{\infty} \frac{1}{m!} = \sum_{m=1}^{\infty} \frac{1}{(m-1)!} + \sum_{m=0}^{\infty} \frac{1}{m!} = e + e = 2e \]

2. The second sum:

\[ 2 \sum_{m=0}^{\infty} \frac{1}{m!} = 2e \]

### Step 6: Combine the Results

Combining both results, we get:

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

### Final Answer

Thus, the sum of the series is:

\[ \boxed{4e} \]