The 6th term of expansion of `(x-1/x)^(10)` is

Options

A. `.^(10)c_(6)x^(6)`
B. `.^(10)c_(5)`
C. `(-(10)c_(5))`
D. `(-^(10)c_(6)x^(6))`

Answer

To find the 6th term of the expansion of \((x - \frac{1}{x})^{10}\), we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (x + y)^n = \sum_{r=0}^{n} \binom{n}{r} x^{n-r} y^r \]

In our case, we have \(x\) and \(y = -\frac{1}{x}\), and \(n = 10\).

### Step 1: Identify the term we need
We need to find the 6th term of the expansion, which corresponds to \(T_{r+1}\) where \(r = 5\) (since \(T_{r+1} = T_6\)).

### Step 2: Write the formula for the term
The general term \(T_r\) in the expansion is given by:

\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]

Substituting \(n = 10\) and \(r = 5\):

\[ T_6 = \binom{10}{5} x^{10-5} \left(-\frac{1}{x}\right)^5 \]

### Step 3: Simplify the expression
Now we can simplify the expression:

\[ T_6 = \binom{10}{5} x^5 \left(-\frac{1}{x}\right)^5 \]

This can be rewritten as:

\[ T_6 = \binom{10}{5} x^5 \cdot \frac{(-1)^5}{x^5} \]

### Step 4: Combine the terms
Since \(x^5\) in the numerator and denominator cancels out, we have:

\[ T_6 = \binom{10}{5} \cdot (-1) \]

### Step 5: Calculate \(\binom{10}{5}\)
Now we calculate \(\binom{10}{5}\):

\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]

### Step 6: Final result
Thus, we have:

\[ T_6 = -252 \]

So, the 6th term of the expansion of \((x - \frac{1}{x})^{10}\) is \(-252\).

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