`(1.3)/(1!) +(2.4)/(2!)+(3.5)/(3!) + (4.6)/(4!) + .......oo = .......`

Options

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

Answer

To solve the series \( S = \frac{1 \cdot 3}{1!} + \frac{2 \cdot 4}{2!} + \frac{3 \cdot 5}{3!} + \frac{4 \cdot 6}{4!} + \ldots \), we can express the general term of the series and simplify it step by step.

### Step 1: Identify the general term
The general term of the series can be expressed as:
\[ T_n = \frac{n(n + 2)}{n!} \]
This is because for each term, the numerator is \( n \) multiplied by \( n + 2 \).

### Step 2: Expand the general term
We can expand \( T_n \):
\[ T_n = \frac{n^2 + 2n}{n!} \]
This can be split into two separate fractions:
\[ T_n = \frac{n^2}{n!} + \frac{2n}{n!} \]

### Step 3: Simplify the fractions
Now, we can simplify each term:
\[ T_n = \frac{n^2}{n!} + \frac{2}{(n-1)!} \]
Here, \( \frac{n^2}{n!} = \frac{n}{(n-1)!} \).

Thus, we can rewrite \( T_n \) as:
\[ T_n = \frac{n}{(n-1)!} + \frac{2}{(n-1)!} = \frac{n + 2}{(n-1)!} \]

### Step 4: Write the series
Now we can write the series \( S \):
\[ S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \left( \frac{n + 2}{(n-1)!} \right) \]

### Step 5: Split the summation
We can split the summation into two parts:
\[ S = \sum_{n=1}^{\infty} \frac{n}{(n-1)!} + \sum_{n=1}^{\infty} \frac{2}{(n-1)!} \]

### Step 6: Change the index of summation
For the first summation, let \( k = n - 1 \):
\[ \sum_{n=1}^{\infty} \frac{n}{(n-1)!} = \sum_{k=0}^{\infty} \frac{k + 1}{k!} = \sum_{k=0}^{\infty} \frac{k}{k!} + \sum_{k=0}^{\infty} \frac{1}{k!} \]
The first part \( \sum_{k=0}^{\infty} \frac{k}{k!} = e \) and the second part \( \sum_{k=0}^{\infty} \frac{1}{k!} = e \).

Thus, the first summation becomes:
\[ \sum_{n=1}^{\infty} \frac{n}{(n-1)!} = e + e = 2e \]

For the second summation:
\[ \sum_{n=1}^{\infty} \frac{2}{(n-1)!} = 2 \sum_{k=0}^{\infty} \frac{1}{k!} = 2e \]

### Step 7: Combine the results
Now, combine both parts:
\[ S = 2e + 2e = 4e \]

### Final Answer
Thus, the value of the series is:
\[ \boxed{4e} \]