The coefficent of `x^(n)` in the series
`1+(a+bx)/(1!)+(a+bx)^(2)/(2!)+(a+bx)^(3)/(3!)`+…is

Options

A. `(b^(n))/(n!)`
B. `e^(b)(b^(n))/(n!)`
C. `e^(a)(b^(n))/(m!)`
D. `(e^(b)a^(n))/(n!)`

Answer

To find the coefficient of \( x^n \) in the series

\[ 1 + \frac{(a + bx)}{1!} + \frac{(a + bx)^2}{2!} + \frac{(a + bx)^3}{3!} + \ldots \]

we can follow these steps:

### Step 1: Identify the General Term
The general term of the series can be expressed as:

\[ \frac{(a + bx)^k}{k!} \]

where \( k \) starts from 0 and goes to infinity.

### Step 2: Expand the General Term
We need to expand \( (a + bx)^k \) using the binomial theorem:

\[ (a + bx)^k = \sum_{j=0}^{k} \binom{k}{j} a^{k-j} (bx)^j = \sum_{j=0}^{k} \binom{k}{j} a^{k-j} b^j x^j \]

### Step 3: Find the Coefficient of \( x^n \)
To find the coefficient of \( x^n \) in the series, we look for the terms where \( j = n \):

\[ \frac{(a + bx)^k}{k!} \text{ contributes } \frac{\binom{k}{n} a^{k-n} b^n}{k!} \]

### Step 4: Sum Over All \( k \)
Now, we need to sum this expression for \( k \) from \( n \) to \( \infty \):

\[ \sum_{k=n}^{\infty} \frac{\binom{k}{n} a^{k-n} b^n}{k!} \]

### Step 5: Simplify the Expression
Using the identity \( \binom{k}{n} = \frac{k!}{n!(k-n)!} \), we can rewrite the sum:

\[ = b^n \sum_{k=n}^{\infty} \frac{a^{k-n}}{(k-n)! n!} \]

Letting \( m = k - n \), we can change the index of summation:

\[ = b^n \frac{1}{n!} \sum_{m=0}^{\infty} \frac{a^m}{m!} = b^n \frac{1}{n!} e^a \]

### Final Result
Thus, the coefficient of \( x^n \) in the series is:

\[ \frac{b^n e^a}{n!} \]