The sum `1+(1+a)/(2!) +(1+a+a^(2))/(3!)+....oo` is equal to

Options

A. `e^(a)`
B. `(e^(a)-e)/(a-1)`
C. `(a-1)e^(a)`
D. `(a+1)e^(a)`

Answer

To solve the series \( S = 1 + \frac{1+a}{2!} + \frac{1+a+a^2}{3!} + \ldots \) up to infinity, we can break down the series into manageable parts.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term of the series can be expressed as:
\[ T_n = \frac{1 + a + a^2 + \ldots + a^{n-1}}{n!} \]
where \( n \) starts from 1.

2. **Sum of the Geometric Series**:
The numerator \( 1 + a + a^2 + \ldots + a^{n-1} \) is a geometric series with first term 1 and common ratio \( a \). The sum of the first \( n \) terms of a geometric series is given by:
\[ S_n = \frac{1 - a^n}{1 - a} \quad \text{(for } a \neq 1\text{)} \]
Therefore, we can rewrite \( T_n \):
\[ T_n = \frac{1 - a^n}{(1 - a)n!} \]

3. **Express the Series**:
Now, substituting \( T_n \) back into the series:
\[ S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{1 - a^n}{(1 - a)n!} \]
This can be separated into two series:
\[ S = \frac{1}{1 - a} \left( \sum_{n=1}^{\infty} \frac{1}{n!} - \sum_{n=1}^{\infty} \frac{a^n}{n!} \right) \]

4. **Recognize the Series**:
The first series \( \sum_{n=1}^{\infty} \frac{1}{n!} = e - 1 \) (since the series for \( e^x \) starts from \( n=0 \)).
The second series \( \sum_{n=1}^{\infty} \frac{a^n}{n!} = e^a - 1 \).

5. **Combine the Results**:
Substituting these results back into the expression for \( S \):
\[ S = \frac{1}{1 - a} \left( (e - 1) - (e^a - 1) \right) \]
Simplifying this gives:
\[ S = \frac{1}{1 - a} (e - e^a) \]

6. **Final Result**:
Thus, the sum of the series is:
\[ S = \frac{e - e^a}{1 - a} \]