`1+3/(1!) +5/(2!) + (7)/(3!) + ......oo=`

Options

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

Answer

To solve the series \( S = 1 + \frac{3}{1!} + \frac{5}{2!} + \frac{7}{3!} + \ldots \), we can follow these steps:

### Step 1: Rewrite the series
We can observe that the series can be rewritten by separating the terms in the numerator:
\[ S = 1 + \left(2 + 1\right) \frac{1}{1!} + \left(4 + 1\right) \frac{1}{2!} + \left(6 + 1\right) \frac{1}{3!} + \ldots \]
This gives us:
\[ S = 1 + \frac{2}{1!} + \frac{4}{2!} + \frac{6}{3!} + \ldots + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots \]

### Step 2: Separate the series
Now we can separate the series into two parts:
\[ S = \left(1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots\right) + \left(\frac{2}{1!} + \frac{4}{2!} + \frac{6}{3!} + \ldots\right) \]

### Step 3: Identify the first series
The first series is the expansion of \( e^x \) at \( x = 1 \):
\[ 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots = e \]

### Step 4: Identify the second series
The second series can be factored:
\[ \frac{2}{1!} + \frac{4}{2!} + \frac{6}{3!} + \ldots = 2\left(\frac{1}{1!} + \frac{2}{2!} + \frac{3}{3!} + \ldots\right) \]
The series inside the parentheses is known to be \( e \) as well:
\[ \frac{1}{1!} + \frac{2}{2!} + \frac{3}{3!} + \ldots = e \]
Thus, we have:
\[ \frac{2}{1!} + \frac{4}{2!} + \frac{6}{3!} + \ldots = 2e \]

### Step 5: Combine the results
Now we can combine both parts:
\[ S = e + 2e = 3e \]

### Final Answer
Thus, the value of the series is:
\[ S = 3e \]