The coefficient of `x^(n)` in the expansion of `(1+x)(1-x)^(n)` is

Options

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

Answer

To find the coefficient of \( x^n \) in the expansion of \( (1+x)(1-x)^n \), we can follow these steps:

### Step 1: Expand \( (1-x)^n \)
Using the Binomial Theorem, we can expand \( (1-x)^n \):
\[ (1-x)^n = \sum_{k=0}^{n} \binom{n}{k} (-x)^k = \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^k \]

### Step 2: Multiply by \( (1+x) \)
Now, we multiply the expansion of \( (1-x)^n \) by \( (1+x) \):
\[ (1+x)(1-x)^n = (1+x) \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^k \]
Distributing \( (1+x) \):
\[ = \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^k + \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^{k+1} \]

### Step 3: Combine the two sums
The first sum gives us the coefficients for \( x^k \), and the second sum gives us the coefficients for \( x^{k+1} \). To find the coefficient of \( x^n \), we need to consider both sums:
- From the first sum, the coefficient of \( x^n \) is \( \binom{n}{n} (-1)^n = (-1)^n \).
- From the second sum, the coefficient of \( x^n \) comes from \( x^{n-1} \), which corresponds to \( k = n-1 \):
\[ \binom{n}{n-1} (-1)^{n-1} = n(-1)^{n-1} \]

### Step 4: Combine coefficients
Now, we combine the coefficients from both contributions:
\[ \text{Coefficient of } x^n = (-1)^n + n(-1)^{n-1} \]

### Step 5: Simplify the expression
Factoring out \( (-1)^{n-1} \):
\[ \text{Coefficient of } x^n = (-1)^{n-1} \left( -1 + n \right) = (-1)^{n-1} (n - 1) \]

### Final Answer
Thus, the coefficient of \( x^n \) in the expansion of \( (1+x)(1-x)^n \) is:
\[ (-1)^{n-1} (n - 1) \]