The coefficient of `x^(n)` in the expansion of
`(1+1/(1!)x+1/(2!)x^(2)…+1/(n!)x^(n))^(2)`

Options

A. `(2^(n))/(n!)`
B. `(2^(n))/n`
C. `n!`
D. `1/(n!)`

Answer

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

\[ \left(1 + \frac{1}{1!}x + \frac{1}{2!}x^2 + \ldots + \frac{1}{n!}x^n\right)^2, \]

we can follow these steps:

### Step 1: Recognize the Inner Series
The expression inside the parentheses is the Taylor series expansion of \( e^x \) truncated at \( n \):

\[ 1 + \frac{1}{1!}x + \frac{1}{2!}x^2 + \ldots + \frac{1}{n!}x^n. \]

This series can be represented as:

\[ f(x) = \sum_{k=0}^{n} \frac{1}{k!} x^k. \]

### Step 2: Square the Series
We need to find the square of this series:

\[ f(x)^2 = \left(\sum_{k=0}^{n} \frac{1}{k!} x^k\right)^2. \]

### Step 3: Use the Cauchy Product Formula
To find the coefficient of \( x^n \) in \( f(x)^2 \), we can use the Cauchy product formula. The coefficient of \( x^n \) is given by:

\[ \sum_{k=0}^{n} a_k a_{n-k}, \]

where \( a_k = \frac{1}{k!} \). Thus, we have:

\[ \text{Coefficient of } x^n = \sum_{k=0}^{n} \frac{1}{k!} \cdot \frac{1}{(n-k)!} = \sum_{k=0}^{n} \frac{1}{k!(n-k)!}. \]

### Step 4: Recognize the Binomial Coefficient
The sum can be recognized as the binomial coefficient:

\[ \sum_{k=0}^{n} \frac{1}{k!(n-k)!} = \frac{1}{n!} \sum_{k=0}^{n} \binom{n}{k} = \frac{2^n}{n!}. \]

### Step 5: Final Coefficient Calculation
Thus, the coefficient of \( x^n \) in \( f(x)^2 \) is:

\[ \frac{2^n}{n!}. \]

### Conclusion
Therefore, the coefficient of \( x^n \) in the expansion of

\[ \left(1 + \frac{1}{1!}x + \frac{1}{2!}x^2 + \ldots + \frac{1}{n!}x^n\right)^2 \]

is

\[ \frac{2^n}{n!}. \]