Which term is independent of 'x' in the expansion of `(2x^2 + 1/x)^12` ? Also , find its value .

Answer

To find the term independent of 'x' in the expansion of \((2x^2 + \frac{1}{x})^{12}\), we will use the Binomial Theorem.

### Step-by-Step Solution:

1. **Identify the General Term:**
The general term \(T_k\) in the expansion of \((a + b)^n\) is given by:
\[ T_k = \binom{n}{k} a^{n-k} b^k \]
In our case, \(a = 2x^2\), \(b = \frac{1}{x}\), and \(n = 12\). Thus, the general term becomes:
\[ T_k = \binom{12}{k} (2x^2)^{12-k} \left(\frac{1}{x}\right)^k \]

2. **Simplify the General Term:**
Now, we simplify \(T_k\):
\[ T_k = \binom{12}{k} (2^{12-k} (x^2)^{12-k}) \left(\frac{1}{x}\right)^k \]
\[ T_k = \binom{12}{k} 2^{12-k} x^{2(12-k)} \cdot x^{-k} \]
\[ T_k = \binom{12}{k} 2^{12-k} x^{24 - 2k - k} \]
\[ T_k = \binom{12}{k} 2^{12-k} x^{24 - 3k} \]

3. **Find the Term Independent of 'x':**
We need the exponent of \(x\) to be zero for the term to be independent of \(x\):
\[ 24 - 3k = 0 \]
Solving for \(k\):
\[ 3k = 24 \implies k = 8 \]

4. **Substitute \(k\) Back to Find the Value:**
Now we substitute \(k = 8\) back into the general term to find the value:
\[ T_8 = \binom{12}{8} 2^{12-8} x^{24 - 3 \cdot 8} \]
\[ T_8 = \binom{12}{8} 2^4 x^{24 - 24} \]
\[ T_8 = \binom{12}{8} 2^4 \]
\[ T_8 = \binom{12}{4} 2^4 \quad (\text{since } \binom{n}{k} = \binom{n}{n-k}) \]
\[ T_8 = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} \cdot 16 \]
\[ T_8 = 495 \cdot 16 = 7920 \]

### Final Answer:
The term independent of \(x\) in the expansion of \((2x^2 + \frac{1}{x})^{12}\) is \(7920\).