The coefficient of `x^9` in the expansion of `(1-5x)/(1+x)` is

Options

A. `-6`
B. 9
C. `-9`
D. 6

Answer

To find the coefficient of \(x^9\) in the expansion of \(\frac{1 - 5x}{1 + x}\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \frac{1 - 5x}{1 + x} \]
This can be rewritten as:
\[ (1 - 5x)(1 + x)^{-1} \]

### Step 2: Expand \((1 + x)^{-1}\)
Using the binomial series expansion, we know that:
\[ (1 + x)^{-1} = \sum_{r=0}^{\infty} (-1)^r x^r = 1 - x + x^2 - x^3 + \ldots \]
Thus, we can express our function as:
\[ (1 - 5x)(1 - x + x^2 - x^3 + \ldots) \]

### Step 3: Distribute \(1 - 5x\) into the series
Now we distribute \(1 - 5x\) into the series:
\[ 1 \cdot (1 - x + x^2 - x^3 + \ldots) - 5x \cdot (1 - x + x^2 - x^3 + \ldots) \]
This gives us:
\[ (1 - x + x^2 - x^3 + \ldots) - (5x - 5x^2 + 5x^3 - 5x^4 + \ldots) \]

### Step 4: Combine the series
Combining the two series, we have:
\[ 1 - x + x^2 - x^3 + \ldots - 5x + 5x^2 - 5x^3 + \ldots \]
This simplifies to:
\[ 1 - 6x + 6x^2 - 6x^3 + \ldots \]

### Step 5: Find the coefficient of \(x^9\)
To find the coefficient of \(x^9\), we look at the terms:
- The \(1\) contributes nothing to \(x^9\).
- The \(-6x\) contributes nothing to \(x^9\).
- The \(6x^2\) contributes nothing to \(x^9\).
- The \(-6x^3\) contributes nothing to \(x^9\).
- Continuing this way, we find that the coefficient of \(x^9\) comes from the term \(-5x^{10}\) when we consider the multiplication of \(5x\) with the \(x^8\) term from the series expansion.

### Step 6: Calculate the coefficient
The coefficient of \(x^9\) from \(5x \cdot x^8\) (where \(x^8\) comes from \((-1)^8 x^8\)) is:
\[ -5 \cdot (-1)^8 = -5 \]
Thus, the total coefficient of \(x^9\) is:
\[ -1 - 5 = -6 \]

### Final Answer
The coefficient of \(x^9\) in the expansion of \(\frac{1 - 5x}{1 + x}\) is \(-6\).