Coefficient of `x^(n)` in the expansion of `((1+x)^(n))/(1-x)`

Options

A. 4n
B. `2^(n)`
C. `n^(2)`
D. `(n(n +1))/(2)`

Answer

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

### Step 1: Rewrite the expression
We start with the expression:
\[ \frac{(1+x)^n}{1-x} \]
This can be rewritten using the formula for the geometric series:
\[ \frac{1}{1-x} = \sum_{k=0}^{\infty} x^k \]
Thus, we have:
\[ \frac{(1+x)^n}{1-x} = (1+x)^n \sum_{k=0}^{\infty} x^k \]

### Step 2: Expand \( (1+x)^n \)
Using the Binomial Theorem, we can expand \( (1+x)^n \):
\[ (1+x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r \]
where \( \binom{n}{r} \) is the binomial coefficient.

### Step 3: Combine the expansions
Now, we combine the two expansions:
\[ \frac{(1+x)^n}{1-x} = \left( \sum_{r=0}^{n} \binom{n}{r} x^r \right) \left( \sum_{k=0}^{\infty} x^k \right) \]
To find the coefficient of \( x^n \) in this product, we need to find all pairs \( (r, k) \) such that \( r + k = n \).

### Step 4: Identify the relevant terms
The pairs \( (r, k) \) that satisfy \( r + k = n \) can be expressed as:
- When \( r = n \), \( k = 0 \): contributes \( \binom{n}{n} = 1 \)
- When \( r = n-1 \), \( k = 1 \): contributes \( \binom{n}{n-1} = n \)
- When \( r = n-2 \), \( k = 2 \): contributes \( \binom{n}{n-2} = \frac{n(n-1)}{2} \)
- ...
- When \( r = 0 \), \( k = n \): contributes \( \binom{n}{0} = 1 \)

### Step 5: Sum the coefficients
The coefficient of \( x^n \) is the sum of all these contributions:
\[ \text{Coefficient of } x^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} \]
This sum is known to equal \( 2^n \) (the sum of the coefficients of the binomial expansion).

### Final Answer
Thus, the coefficient of \( x^n \) in the expansion of \( \frac{(1+x)^n}{1-x} \) is:
\[ \boxed{2^n} \]