Constant term in the expansion of ` ( x- (1)/(x)) ^(10) ` is

Options

A. `152`
B. ` -152`
C. ` -252`
D. ` 252`

Answer

To find the constant term in the expansion of \( (x - \frac{1}{x})^{10} \), we can follow these steps:

### Step 1: Identify the general term
The general term (T) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = x \), \( b = -\frac{1}{x} \), and \( n = 10 \). Thus, the general term becomes:
\[ T_{r+1} = \binom{10}{r} x^{10-r} \left(-\frac{1}{x}\right)^r \]

### Step 2: Simplify the general term
Simplifying the general term, we have:
\[ T_{r+1} = \binom{10}{r} x^{10-r} \cdot (-1)^r \cdot x^{-r} = \binom{10}{r} (-1)^r x^{10 - 2r} \]

### Step 3: Find the constant term
To find the constant term, we need to set the exponent of \( x \) to zero:
\[ 10 - 2r = 0 \]
Solving for \( r \):
\[ 10 = 2r \implies r = 5 \]

### Step 4: Substitute \( r \) back into the general term
Now, substituting \( r = 5 \) into the general term:
\[ T_{5+1} = \binom{10}{5} (-1)^5 \]

### Step 5: Calculate \( \binom{10}{5} \)
Calculating \( \binom{10}{5} \):
\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]

### Step 6: Include the sign from \( (-1)^5 \)
Since \( (-1)^5 = -1 \), we have:
\[ T_{6} = 252 \cdot (-1) = -252 \]

### Conclusion
Thus, the constant term in the expansion of \( (x - \frac{1}{x})^{10} \) is:
\[ \boxed{-252} \]