If the constant term of the binomial expansion `(2x -1/x)^n` is `-160`, then n is equal to -

Options

A. 4
B. 6
C. 8
D. 10

Answer

To find the value of \( n \) such that the constant term of the binomial expansion \( (2x - \frac{1}{x})^n \) is equal to \(-160\), we will follow these steps:

### Step 1: Identify the General Term
The general term \( T_r \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = 2x \) and \( b = -\frac{1}{x} \). Therefore, the general term becomes:
\[ T_r = \binom{n}{r} (2x)^{n-r} \left(-\frac{1}{x}\right)^r \]

### Step 2: Simplify the General Term
Now, simplifying \( T_r \):
\[ T_r = \binom{n}{r} (2^{n-r} x^{n-r}) \left(-1^r \cdot x^{-r}\right) \]
This simplifies to:
\[ T_r = \binom{n}{r} 2^{n-r} (-1)^r x^{n - 2r} \]

### Step 3: Find the Constant Term
For the term to be constant, the exponent of \( x \) must be zero:
\[ n - 2r = 0 \implies r = \frac{n}{2} \]
This means \( n \) must be even.

### Step 4: Substitute \( r \) into the General Term
Substituting \( r = \frac{n}{2} \) into the general term:
\[ T_{\frac{n}{2}} = \binom{n}{\frac{n}{2}} 2^{n - \frac{n}{2}} (-1)^{\frac{n}{2}} = \binom{n}{\frac{n}{2}} 2^{\frac{n}{2}} (-1)^{\frac{n}{2}} \]

### Step 5: Set the Constant Term Equal to -160
We know this term equals \(-160\):
\[ \binom{n}{\frac{n}{2}} 2^{\frac{n}{2}} (-1)^{\frac{n}{2}} = -160 \]
This implies:
\[ -\binom{n}{\frac{n}{2}} 2^{\frac{n}{2}} = -160 \implies \binom{n}{\frac{n}{2}} 2^{\frac{n}{2}} = 160 \]

### Step 6: Factor 160
Now, we can express 160 in terms of its factors:
\[ 160 = 2^5 \times 5 \]

### Step 7: Solve for \( n \)
We can try different even values for \( n \) to find \( \binom{n}{\frac{n}{2}} 2^{\frac{n}{2}} = 160 \).

1. **For \( n = 4 \)**:
\[ \binom{4}{2} 2^{2} = 6 \times 4 = 24 \quad (\text{not } 160) \]

2. **For \( n = 6 \)**:
\[ \binom{6}{3} 2^{3} = 20 \times 8 = 160 \quad (\text{this works!}) \]

3. **For \( n = 8 \)**:
\[ \binom{8}{4} 2^{4} = 70 \times 16 = 1120 \quad (\text{not } 160) \]

4. **For \( n = 10 \)**:
\[ \binom{10}{5} 2^{5} = 252 \times 32 = 8064 \quad (\text{not } 160) \]

### Conclusion
The only value of \( n \) that satisfies the condition is:
\[ \boxed{6} \]