The coefficient of `x^(5)` in the expansion of
`(1+x^(2))/(1 +x) ,|x| lt 1 `, is

Options

A. -1
B. 2
C. 0
D. -2

Answer

To find the coefficient of \( x^5 \) in the expansion of \( \frac{1 + x^2}{1 + x} \), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ \frac{1 + x^2}{1 + x} \]
This can be rewritten as:
\[ (1 + x^2)(1 + x)^{-1} \]

### Step 2: Expand \( (1 + x)^{-1} \)
Using the binomial series expansion for \( (1 + x)^{-1} \), we have:
\[ (1 + x)^{-1} = 1 - x + x^2 - x^3 + x^4 - x^5 + \ldots \]
This series converges for \( |x| < 1 \).

### Step 3: Multiply the Expansions
Now we multiply \( (1 + x^2) \) with the expansion of \( (1 + x)^{-1} \):
\[ (1 + x^2)(1 - x + x^2 - x^3 + x^4 - x^5 + \ldots) \]
Distributing \( 1 + x^2 \) gives:
\[ 1(1 - x + x^2 - x^3 + x^4 - x^5 + \ldots) + x^2(1 - x + x^2 - x^3 + x^4 - x^5 + \ldots) \]

### Step 4: Identify Terms Contributing to \( x^5 \)
Now we need to find the terms that contribute to \( x^5 \):
1. From \( 1 \cdot (-x^5) \), we get \( -1 \).
2. From \( x^2 \cdot x^3 \), we get \( +1 \).

### Step 5: Combine the Coefficients
Adding the contributions:
\[ -1 + 1 = 0 \]

### Conclusion
Thus, the coefficient of \( x^5 \) in the expansion of \( \frac{1 + x^2}{1 + x} \) is:
\[ \boxed{-2} \]