The sum of the series
`1+(1+a)/(2!)+(1+a+a^(2))/(3!)+(1+a+a^(2)+a^(3))/(4!)`+…..is

Options

A. `(e^(a)-e)/(a-1)`
B. `(e^(a)-e)/(a+1)`
C. `(e^(2a)+1)/(a-1)`
D. `(e^(a)+e)/(a-1)`

Answer

To find the sum of the series

\[ S = 1 + \frac{1+a}{2!} + \frac{1+a+a^2}{3!} + \frac{1+a+a^2+a^3}{4!} + \ldots \]

we can express the general term of the series. The \(n\)-th term can be written as:

\[ T_n = \frac{1 + a + a^2 + \ldots + a^{n-1}}{n!} \]

The expression \(1 + a + a^2 + \ldots + a^{n-1}\) is a geometric series with \(n\) terms, first term \(1\), and common ratio \(a\). The sum of this geometric series is given by:

\[ 1 + a + a^2 + \ldots + a^{n-1} = \frac{1 - a^n}{1 - a} \quad \text{(for } a \neq 1\text{)} \]

Thus, we can rewrite \(T_n\) as:

\[ T_n = \frac{1 - a^n}{(1 - a)n!} \]

Now, substituting this back into the series, we have:

\[ S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{1 - a^n}{(1 - a)n!} \]

This can be separated into two sums:

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

The first sum, \(\sum_{n=1}^{\infty} \frac{1}{n!}\), is the Taylor series expansion for \(e\), so:

\[ \sum_{n=1}^{\infty} \frac{1}{n!} = e - 1 \]

The second sum, \(\sum_{n=1}^{\infty} \frac{a^n}{n!}\), is the Taylor series expansion for \(e^a\), so:

\[ \sum_{n=1}^{\infty} \frac{a^n}{n!} = e^a - 1 \]

Substituting these results back into the expression for \(S\):

\[ S = \frac{1}{1 - a} \left( (e - 1) - (e^a - 1) \right) \]

Simplifying this gives:

\[ S = \frac{1}{1 - a} \left( e - e^a \right) \]

Thus, the final result for the sum of the series is:

\[ S = \frac{e - e^a}{1 - a} \]