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 becomes:
\[ T_{r+1} = \binom{11}{r} (2x^4)^{11-r} \left(-\frac{1}{3x^7}\right)^r \]

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

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

### Step 4: Substitute \( r \) into the General Term
Now we substitute \( r = 4 \) back into the general term to find the constant term:
\[ T_{5} = \binom{11}{4} 2^{11-4} (-1)^4 \frac{1}{3^4} \]
Calculating each part:
\[ \binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330 \]
\[ 2^{11-4} = 2^7 = 128 \]
\[ (-1)^4 = 1 \]
\[ 3^4 = 81 \]

### Step 5: Combine the Values
Now we can combine these values to find the constant term:
\[ T_{5} = 330 \cdot 128 \cdot \frac{1}{81} \]
Calculating:
\[ T_{5} = \frac{330 \cdot 128}{81} \]
Calculating \( 330 \cdot 128 = 42240 \):
\[ T_{5} = \frac{42240}{81} \]

### Final Answer
Thus, the constant term in the expansion of \( (2x^4 - \frac{1}{3x^7})^{11} \) is:
\[ \frac{42240}{81} \]