Question

The coefficient of \( x^2 \) in the binomial expansion of \( (2x^2 + \frac{1}{x})^{10 \) is:

• A. 3360
• B. 2360
• C. 3260
• D. 3380

Hint:

To find the coefficient of $x^k$ in $(ax^p + bx^q)^n$, use the formula $r = \frac{np - k}{p - q}$. Here, $r = \frac{10(2) - 2}{2 - (-1)} = \frac{18}{3} = 6$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We use the general term formula for a binomial expansion $(a+b)^n$, which is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. We need to find the value of $r$ that results in the power of $x$ being 2.

Step 2: Key Formula or Approach:
1. General term: $T_{r+1} = \binom{10}{r} (2x^2)^{10-r} (x^{-1})^r$. 2. Simplify the powers of $x$ and set the exponent equal to 2.

Step 3: Detailed Explanation:
1. Express $T_{r+1}$: \[ T_{r+1} = \binom{10}{r} 2^{10-r} \cdot x^{2(10-r)} \cdot x^{-r} = \binom{10}{r} 2^{10-r} \cdot x^{20-2r-r} = \binom{10}{r} 2^{10-r} \cdot x^{20-3r} \] 2. For $x^2$, we need $20 - 3r = 2$: \[ 3r = 18 \implies r = 6 \] 3. Calculate the coefficient for $r=6$: \[ \text{Coeff} = \binom{10}{6} 2^{10-6} = \binom{10}{4} 2^4 \] \[ \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \] \[ \text{Coeff} = 210 \times 16 = 3360 \]

Step 4: Final Answer:
The coefficient of $x^2$ is 3360.