Question

The constant term in the expansion of $(2x^{2}-\frac{1}{x^{2}})^{6}$ is ________.

• A. -160
• B. 160
• C. -180
• D. 180
• E. 0

Hint:

Constant term means the $x^0$ term.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The "constant term" in a polynomial expansion is the term that does not contain the variable \(x\), which is equivalent to the coefficient of \(x^0\). We need to find which term in the binomial expansion of \((x - \frac{2}{x})^6\) results in \(x^0\).
Step 2: Key Formula or Approach
The general term in the expansion of \((A+B)^n\) is \(T_{r+1} = {}^nC_r A^{n-r} B^r\). We will find the general term for the given expression, determine the value of \(r\) that makes the power of \(x\) equal to zero, and then calculate the coefficient for that value of \(r\).
Step 3: Detailed Explanation
1. Identify A, B, and n.
For the expansion of \((x - \frac{2}{x})^6\):
\(A = x\)
\(B = -\frac{2}{x} = -2x^{-1}\)
\(n = 6\)
2. Write down the general term.
\[ T_{r+1} = {}^6C_r (x)^{6-r} \left(-2x^{-1}\right)^r \] 3. Separate the constant and variable parts.
\[ T_{r+1} = {}^6C_r (-2)^r (x)^{6-r} (x^{-1})^r \] 4. Simplify the power of x.
\[ x^{6-r} \cdot x^{-r} = x^{6-r-r} = x^{6-2r} \] 5. Find the value of r for the constant term.
For the constant term, the power of \(x\) must be 0.
\[ 6 - 2r = 0 \] \[ 2r = 6 \] \[ r = 3 \] This means the constant term is the \((3+1)\)th term, i.e., the 4th term.
6. Calculate the coefficient for r = 3.
Substitute \(r=3\) back into the coefficient part of the general term, which is \({}^6C_r (-2)^r\).
Coefficient = \({}^6C_3 (-2)^3\)
First, calculate \({}^6C_3\):
\[ {}^6C_3 = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \] Next, calculate \((-2)^3\):
\[ (-2)^3 = -8 \] Now, multiply these values:
Constant Term = \(20 \times (-8) = -160\)
Step 4: Final Answer
The constant term in the expansion is -160.