Question

The constant term in the expansion of \( \left(x^2 - \frac{2}{x}\right)^6 \) is

• A. \(60\)
• B. \(180\)
• C. \(240\)
• D. \(360\)
• E. \(420\)

Hint:

Always equate exponent of variable to zero to find constant term.

Answer 1

The Correct Option is C

Concept: The constant term occurs when the power of \(x\) becomes zero in the general term.

Step 1: Write general term using binomial theorem. \[ T_k = \binom{6}{k}(x^2)^{6-k}\left(-\frac{2}{x}\right)^k \]

Step 2: Simplify powers of \(x\). \[ (x^2)^{6-k} = x^{12 - 2k} \] \[ \left(\frac{1}{x}\right)^k = x^{-k} \] So total power: \[ x^{12 - 2k} \cdot x^{-k} = x^{12 - 3k} \] Thus: \[ T_k = \binom{6}{k}(-2)^k x^{12-3k} \]

Step 3: Find when power is zero. For constant term: \[ 12 - 3k = 0 \] \[ 3k = 12 \Rightarrow k = 4 \]

Step 4: Substitute \(k=4\). \[ T_4 = \binom{6}{4}(-2)^4 \] \[ = 15 \times 16 \] \[ = 240 \]