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

Options

A. `n-1`
B. `(-1)^(n)(1-n)`
C. `(-1)^(n-1) (n-1)^(2)`
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
The Binomial Theorem states that:
\[ (1 - x)^n = \sum_{k=0}^{n} \binom{n}{k} (-x)^k = \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^k \]
Thus, the expansion of \( (1-x)^n \) is:
\[ (1-x)^n = \binom{n}{0} - \binom{n}{1} x + \binom{n}{2} x^2 - \binom{n}{3} x^3 + \ldots + (-1)^n \binom{n}{n} x^n \]

### Step 2: Multiply by \( (1+x) \)
Now we multiply the expansion of \( (1-x)^n \) by \( (1+x) \):
\[ (1+x)(1-x)^n = (1+x) \left( \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^k \right) \]
Distributing \( (1+x) \):
\[ = \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^k + x \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^k \]
The first term gives us the coefficients of \( x^k \) directly, while the second term shifts the coefficients by one degree.

### Step 3: Collect the coefficients of \( x^n \)
The coefficient of \( x^n \) in the first sum is \( \binom{n}{n} (-1)^n = (-1)^n \).
In the second sum, the coefficient of \( x^n \) comes from the \( x^{n-1} \) term of \( (1-x)^n \), which is \( \binom{n}{n-1} (-1)^{n-1} = -(-1)^{n-1} n \).

Combining these, the coefficient of \( x^n \) in \( (1+x)(1-x)^n \) is:
\[ (-1)^n + (-1)^{n-1} n = (-1)^n - n(-1)^n = (1-n)(-1)^n \]

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