Coefficient of `x^n` in `(1+x)/(1!) +((1+x)^2)/(2!) + ((1+x)^3)/(3!) + .......=`

Options

A. `e/(n!)`
B. `1/(e.n!)`
C. `e/(n!) -1`
D. `(en)/(n!)`

Answer

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

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

we can follow these steps:

### Step 1: Recognize the Series
The series can be rewritten as:

\[ \sum_{k=1}^{\infty} \frac{(1+x)^k}{k!} \]

This resembles the Taylor series expansion of \( e^{x} \), but we have \( (1+x)^k \) instead.

### Step 2: Rewrite the Series
We can express the series as:

\[ \sum_{k=1}^{\infty} \frac{(1+x)^k}{k!} = e^{1+x} - 1 \]

This is because the series for \( e^{u} \) is \( \sum_{k=0}^{\infty} \frac{u^k}{k!} \), and here \( u = 1+x \).

### Step 3: Simplify the Expression
Thus, we have:

\[ e^{1+x} - 1 = e \cdot e^x - 1 \]

### Step 4: Find the Coefficient of \( x^n \)
Now, we need to find the coefficient of \( x^n \) in \( e \cdot e^x - 1 \). The expansion of \( e^x \) is:

\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]

The coefficient of \( x^n \) in \( e^x \) is \( \frac{1}{n!} \).

### Step 5: Multiply by \( e \)
Thus, the coefficient of \( x^n \) in \( e \cdot e^x \) is:

\[ e \cdot \frac{1}{n!} \]

### Step 6: Final Result
Since we are looking for the coefficient of \( x^n \) in \( e^{1+x} - 1 \), we find that:

\[ \text{Coefficient of } x^n = \frac{e}{n!} \]

### Conclusion
Therefore, the coefficient of \( x^n \) in the given series is:

\[ \frac{e}{n!} \]

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