`1 + (3)/(1!) + (5)/(2!) + (7)/(3!) + (9)/(4!) + ... = `..........

Answer

To solve the series \( S = 1 + \frac{3}{1!} + \frac{5}{2!} + \frac{7}{3!} + \frac{9}{4!} + \ldots \), we can follow these steps:

### Step 1: Identify the General Term
The series can be expressed in terms of a general term. Observing the pattern, we can see that the \( n \)-th term can be written as:
\[ \text{Term}_n = \frac{(2n + 1)}{n!} \]
where \( n \) starts from 0.

### Step 2: Rewrite the Series
Thus, we can rewrite the series as:
\[ S = \sum_{n=0}^{\infty} \frac{(2n + 1)}{n!} \]

### Step 3: Split the Series
We can split the series into two parts:
\[ S = \sum_{n=0}^{\infty} \frac{2n}{n!} + \sum_{n=0}^{\infty} \frac{1}{n!} \]
The first part can be simplified as follows:
\[ \sum_{n=0}^{\infty} \frac{2n}{n!} = 2 \sum_{n=1}^{\infty} \frac{1}{(n-1)!} = 2 \sum_{m=0}^{\infty} \frac{1}{m!} = 2e \]
where we made the substitution \( m = n - 1 \).

### Step 4: Evaluate the Second Part
The second part is:
\[ \sum_{n=0}^{\infty} \frac{1}{n!} = e \]

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

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

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