Find the middle term in the expansion of : `\ (x-1/x)^(10)`

Options

A. 126
B. -126
C. -252
D. 252

Explanation

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

### Step 1: Identify the Binomial Expansion
The binomial expansion of \((a + b)^n\) is given by:
\[ \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, \(a = x\), \(b = -\frac{1}{x}\), and \(n = 10\).

### Step 2: Write the Expansion
Using the binomial theorem, we can write the expansion:
\[ (x - \frac{1}{x})^{10} = \sum_{k=0}^{10} \binom{10}{k} x^{10-k} \left(-\frac{1}{x}\right)^k \]
This simplifies to:
\[ = \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^{10 - k - k} = \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^{10 - 2k} \]

### Step 3: Determine the Middle Term
The total number of terms in the expansion is \(n + 1 = 10 + 1 = 11\). Therefore, the middle term will be the \(\frac{11}{2}\)th term, which is the 6th term (since we round up).

### Step 4: Find the 6th Term
The 6th term corresponds to \(k = 5\) (because we start counting from \(k=0\)):
\[ \text{6th term} = \binom{10}{5} (-1)^5 x^{10 - 2 \cdot 5} \]
Calculating this gives:
\[ = \binom{10}{5} (-1)^5 x^{0} = \binom{10}{5} (-1) \]

### Step 5: Calculate \(\binom{10}{5}\)
\[ \binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \]
Calculating the factorials:
\[ = \frac{10 \times 9 \times 8 \times 7 \times 6}{120} \]
Calculating the numerator:
\[ = 10 \times 9 = 90 \]
\[ 90 \times 8 = 720 \]
\[ 720 \times 7 = 5040 \]
\[ 5040 \times 6 = 30240 \]
Now dividing by \(120\):
\[ \frac{30240}{120} = 252 \]
Thus, \(\binom{10}{5} = 252\).

### Step 6: Final Calculation
Now substituting back:
\[ \text{6th term} = 252 \cdot (-1) = -252 \]

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

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