The coefficient of `x^(n)` in the expansion of `((1+x)^(2))/((1 - x)^(3))`, is

Options

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

Answer

To find the coefficient of \( x^n \) in the expansion of \( \frac{(1+x)^2}{(1-x)^3} \), we can follow these steps:

### Step 1: Expand \( (1+x)^2 \)
Using the binomial theorem, we can expand \( (1+x)^2 \):
\[ (1+x)^2 = 1 + 2x + x^2 \]

### Step 2: Expand \( (1-x)^{-3} \)
Using the binomial series for negative exponents, we expand \( (1-x)^{-3} \):
\[ (1-x)^{-3} = \sum_{k=0}^{\infty} \binom{k+2}{2} x^k \]
This gives us the coefficients for \( x^k \) as \( \binom{k+2}{2} \).

### Step 3: Combine the expansions
Now, we need to combine the two expansions:
\[ \frac{(1+x)^2}{(1-x)^3} = (1 + 2x + x^2) \cdot \left( \sum_{k=0}^{\infty} \binom{k+2}{2} x^k \right) \]

### Step 4: Find the coefficient of \( x^n \)
To find the coefficient of \( x^n \), we consider contributions from each term in \( (1 + 2x + x^2) \):
- From \( 1 \): The coefficient of \( x^n \) is \( \binom{n+2}{2} \).
- From \( 2x \): The coefficient of \( x^{n-1} \) is \( 2 \cdot \binom{(n-1)+2}{2} = 2 \cdot \binom{n+1}{2} \).
- From \( x^2 \): The coefficient of \( x^{n-2} \) is \( \binom{(n-2)+2}{2} = \binom{n}{2} \).

Thus, the total coefficient of \( x^n \) is:
\[ \binom{n+2}{2} + 2 \cdot \binom{n+1}{2} + \binom{n}{2} \]

### Step 5: Simplify the expression
Now we simplify this expression:
1. Recall that \( \binom{n+2}{2} = \frac{(n+2)(n+1)}{2} \)
2. \( \binom{n+1}{2} = \frac{(n+1)n}{2} \)
3. \( \binom{n}{2} = \frac{n(n-1)}{2} \)

Putting it all together:
\[ \text{Total Coefficient} = \frac{(n+2)(n+1)}{2} + 2 \cdot \frac{(n+1)n}{2} + \frac{n(n-1)}{2} \]
\[ = \frac{(n+2)(n+1) + 2(n+1)n + n(n-1)}{2} \]

### Step 6: Combine like terms
Now we combine the terms:
\[ = \frac{n^2 + 3n + 2 + 2n^2 + 2n + n^2 - n}{2} \]
\[ = \frac{4n^2 + 4n + 2}{2} \]
\[ = 2n^2 + 2n + 1 \]

### Final Answer
Thus, the coefficient of \( x^n \) in the expansion of \( \frac{(1+x)^2}{(1-x)^3} \) is:
\[ \boxed{2n^2 + 2n + 1} \]