Find the constant term in the expansion of `(2x^(4)-(1)/(3x^(7)))^(11)`

Explanation

To find the constant term in the expansion of \((2x^{4} - \frac{1}{3x^{7}})^{11}\), we will follow these steps:

### Step 1: Identify the General Term
The general term \(T_{r+1}\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = 2x^{4}\), \(b = -\frac{1}{3x^{7}}\), and \(n = 11\). Therefore, the general term can be written as:
\[ T_{r+1} = \binom{11}{r} (2x^{4})^{11-r} \left(-\frac{1}{3x^{7}}\right)^{r} \]

### Step 2: Simplify the General Term
Now, we simplify \(T_{r+1}\):
\[ T_{r+1} = \binom{11}{r} (2^{11-r} (x^{4})^{11-r}) \left(-\frac{1}{3}\right)^{r} (x^{-7})^{r} \]
This simplifies to:
\[ T_{r+1} = \binom{11}{r} 2^{11-r} (-1)^{r} \frac{1}{3^{r}} x^{4(11-r) - 7r} \]
\[ = \binom{11}{r} 2^{11-r} (-1)^{r} \frac{1}{3^{r}} x^{44 - 4r - 7r} \]
\[ = \binom{11}{r} 2^{11-r} (-1)^{r} \frac{1}{3^{r}} x^{44 - 11r} \]

### Step 3: Find the Constant Term
The constant term is the term where the exponent of \(x\) is zero:
\[ 44 - 11r = 0 \]
Solving for \(r\):
\[ 44 = 11r \implies r = 4 \]

### Step 4: Substitute \(r\) into the General Term
Now, we substitute \(r = 4\) back into the general term:
\[ T_{5} = \binom{11}{4} 2^{11-4} (-1)^{4} \frac{1}{3^{4}} \]
Calculating each part:
\[ = \binom{11}{4} 2^{7} \cdot 1 \cdot \frac{1}{81} \]
\[ = \binom{11}{4} \cdot 128 \cdot \frac{1}{81} \]

### Step 5: Calculate \(\binom{11}{4}\)
\[ \binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330 \]

### Step 6: Final Calculation
Now we can calculate the constant term:
\[ \text{Constant Term} = 330 \cdot 128 \cdot \frac{1}{81} \]
Calculating \(330 \cdot 128\):
\[ 330 \cdot 128 = 42240 \]
Now divide by \(81\):
\[ \text{Constant Term} = \frac{42240}{81} = 520 \]

### Final Answer
The constant term in the expansion is \(520\).