For `|x| lt 1`, the constant term in the expansion of `(1)/((x-1)^(2)(x-2)` is :

Options

A. a) 2
B. b) 1
C. c) 0
D. d) `- 1/2`

Answer

To find the constant term in the expansion of \(\frac{1}{(x-1)^2 (x-2)}\) for \(|x| < 1\), we will follow these steps:

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

### Step 2: Use the Binomial Expansion
We can use the binomial expansion for each term. The binomial expansion for \((1 + u)^n\) is given by:
\[ (1 + u)^n = \sum_{r=0}^{\infty} \binom{n}{r} u^r \]
We will rewrite \((x-1)^{-2}\) and \((x-2)^{-1}\) in a suitable form for expansion.

### Step 3: Rewrite Each Factor
We can express \((x-1)^{-2}\) and \((x-2)^{-1}\) as:
\[ (x-1)^{-2} = \left(-1(1-x)\right)^{-2} = (-1)^{-2} (1-x)^{-2} = (1-x)^{-2} \]
\[ (x-2)^{-1} = \left(-2(1-\frac{x}{2})\right)^{-1} = -\frac{1}{2}(1 - \frac{x}{2})^{-1} \]

### Step 4: Expand Each Factor
Using the binomial series expansion:
1. For \((1-x)^{-2}\):
\[ (1-x)^{-2} = \sum_{r=0}^{\infty} \binom{r+1}{1} x^r = \sum_{r=0}^{\infty} (r+1) x^r \]
2. For \((1 - \frac{x}{2})^{-1}\):
\[ (1 - \frac{x}{2})^{-1} = \sum_{r=0}^{\infty} \left(\frac{x}{2}\right)^r = \sum_{r=0}^{\infty} \frac{x^r}{2^r} \]

### Step 5: Combine the Expansions
Now we need to multiply the two series:
\[ (1-x)^{-2} \cdot (1 - \frac{x}{2})^{-1} = \left(\sum_{r=0}^{\infty} (r+1) x^r\right) \cdot \left(\sum_{s=0}^{\infty} \frac{x^s}{2^s}\right) \]

### Step 6: Find the Constant Term
To find the constant term, we need to consider the coefficient of \(x^0\) in the combined series. This occurs when \(r+s=0\), which means \(r=0\) and \(s=0\):
\[ \text{Constant term} = (1)(\frac{1}{2^0}) = 1 \]

### Step 7: Multiply by the Coefficient from \((x-2)^{-1}\)
Since we have \(-\frac{1}{2}\) from the expansion of \((x-2)^{-1}\):
\[ \text{Constant term} = 1 \cdot -\frac{1}{2} = -\frac{1}{2} \]

### Final Answer
Thus, the constant term in the expansion of \(\frac{1}{(x-1)^2 (x-2)}\) for \(|x| < 1\) is:
\[ \boxed{-\frac{1}{2}} \]