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-by-Step Solution:

1. **Identify the terms in the binomial expansion**:
The general term in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \(a = x\), \(b = -\frac{1}{x}\), and \(n = 10\).

2. **Write the general term**:
For our expression, the \(r^{th}\) term (or \(T_{r+1}\)) will be:
\[ T_{r+1} = \binom{10}{r} x^{10-r} \left(-\frac{1}{x}\right)^r \]
Simplifying this gives:
\[ T_{r+1} = \binom{10}{r} x^{10-r} \cdot \left(-1\right)^r \cdot \frac{1}{x^r} = \binom{10}{r} (-1)^r x^{10 - r - r} = \binom{10}{r} (-1)^r x^{10 - 2r} \]

3. **Find the constant term**:
The constant term occurs when the exponent of \(x\) is zero:
\[ 10 - 2r = 0 \]
Solving for \(r\):
\[ 10 = 2r \implies r = 5 \]

4. **Substitute \(r\) back into the term**:
Now we substitute \(r = 5\) into the general term:
\[ T_{6} = \binom{10}{5} (-1)^5 \]
We know \((-1)^5 = -1\), so:
\[ T_{6} = \binom{10}{5} \cdot (-1) \]

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

6. **Final calculation of the constant term**:
Therefore, the constant term is:
\[ T_{6} = -252 \]

### Conclusion:
The constant term in the expansion of \((x - \frac{1}{x})^{10}\) is \(-252\).

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