Question

The coefficient of $x^2$ in the expansion of $\left( 2x^2 + \frac{1}{x} \right)^{10}, x \neq 0$, is :

• A. 3240
• B. 3360
• C. 3480
• D. 3600

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Use the general term of the binomial expansion $(a + b)^n$, which is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.
: Key Formula or Approach:
Expansion: $\left( 2x^2 + x^{-1} \right)^{10}$.
$T_{r+1} = \binom{10}{r} (2x^2)^{10-r} (x^{-1})^r = \binom{10}{r} 2^{10-r} x^{20-2r} x^{-r} = \binom{10}{r} 2^{10-r} x^{20-3r}$.
Step 2: Detailed Explanation:
We need the coefficient of $x^2$. Therefore, set the exponent of $x$ to 2:
$20 - 3r = 2 \implies 3r = 18 \implies r = 6$.
Substitute $r = 6$ back into the term:
Coefficient $= \binom{10}{6} 2^{10-6} = \binom{10}{4} 2^4$.
$\binom{10}{4} = \frac{10 \cdot 9 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 2 \cdot 1} = 210$.
Coefficient $= 210 \times 16 = 3360$.
Step 3: Final Answer:
The coefficient is 3360.