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:
In the binomial expansion of \((a + b)^n\), the general term \(T_{r+1}\) is given by \(^nC_r \cdot a^{n-r} \cdot b^r\).
We need to find the specific value of \(r\) for which the power of \(x\) equals 2.
Step 2: Key Formula or Approach:
For the expansion of \(\left(2x^2 + \frac{1}{x}\right)^{10}\):
\[ T_{r+1} = ^{10}C_r \cdot (2x^2)^{10-r} \cdot \left(\frac{1}{x}\right)^r \]
\[ T_{r+1} = ^{10}C_r \cdot 2^{10-r} \cdot x^{2(10-r)} \cdot x^{-r} \]
\[ T_{r+1} = ^{10}C_r \cdot 2^{10-r} \cdot x^{20 - 2r - r} \]
\[ T_{r+1} = ^{10}C_r \cdot 2^{10-r} \cdot x^{20 - 3r} \]
Step 3: Detailed Explanation:
To find the coefficient of \(x^2\), set the exponent of \(x\) to 2:
\[ 20 - 3r = 2 \]
\[ 3r = 18 \Rightarrow r = 6 \]
Substitute \(r = 6\) back into the coefficient expression:
\[ \text{Coefficient} = ^{10}C_6 \cdot 2^{10-6} = ^{10}C_4 \cdot 2^4 \]
\[ ^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210 \]
\[ \text{Coefficient} = 210 \times 16 = 3360 \]
Step 4: Final Answer:
The coefficient of \(x^2\) is 3360.