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: Rewrite the expression
We can rewrite the expression as:

\[ (1+x)^2 \cdot (1-x)^{-3}. \]

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

\[ (1+x)^2 = 1 + 2x + x^2. \]

### Step 3: Expand \( (1-x)^{-3} \)
Using the binomial series for negative exponents, we have:

\[ (1-x)^{-3} = \sum_{k=0}^{\infty} \binom{k+2}{2} x^k. \]

This is derived from the general formula for the expansion of \( (1-x)^{-m} \), which is:

\[ (1-x)^{-m} = \sum_{k=0}^{\infty} \binom{m+k-1}{k} x^k. \]

For \( m = 3 \), this gives us \( \binom{k+2}{2} \).

### Step 4: Combine the expansions
Now we need to multiply the two expansions:

\[ (1 + 2x + x^2) \cdot \left( \sum_{k=0}^{\infty} \binom{k+2}{2} x^k \right). \]

### Step 5: Find the coefficient of \( x^n \)
To find the coefficient of \( x^n \), we consider the contributions from each term in \( (1 + 2x + x^2) \):

1. From \( 1 \): The coefficient of \( x^n \) is \( \binom{n+2}{2} \).
2. From \( 2x \): The coefficient of \( x^{n-1} \) is \( 2 \binom{(n-1)+2}{2} = 2 \binom{n+1}{2} \).
3. 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:

\[ \binom{n+2}{2} + 2 \binom{n+1}{2} + \binom{n}{2}. \]

### Step 6: Simplify the expression
Now we can simplify this expression:

1. Recall that \( \binom{n}{2} = \frac{n(n-1)}{2} \), \( \binom{n+1}{2} = \frac{(n+1)n}{2} \), and \( \binom{n+2}{2} = \frac{(n+2)(n+1)}{2} \).

2. Substitute these into our expression:

\[ \frac{(n+2)(n+1)}{2} + 2 \cdot \frac{(n+1)n}{2} + \frac{n(n-1)}{2}. \]

3. 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

\[ \boxed{2n^2 + 2n + 1}. \]