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: Write the general term in the binomial expansion
The general term (the \( r+1 \)th term) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our expression, let \( a = 2x^4 \) and \( b = -\frac{1}{3x^7} \), and \( n = 11 \). Therefore, the \( r+1 \)th term is:
\[ T_{r+1} = \binom{11}{r} (2x^4)^{11-r} \left(-\frac{1}{3x^7}\right)^r \]

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

### Step 3: Find the constant term
For the term to be constant, the exponent of \( x \) must be zero:
\[ 44 - 11r = 0 \]
Solving for \( r \):
\[ 11r = 44 \implies r = 4 \]

### Step 4: Substitute \( r \) back to find the constant term
Now, substitute \( r = 4 \) into the general term:
\[ T_{5} = \binom{11}{4} (-1)^4 \frac{2^{11-4}}{3^4} \]
Calculating this:
\[ T_{5} = \binom{11}{4} \frac{2^7}{3^4} \]
Calculating \( \binom{11}{4} \):
\[ \binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330 \]
Now substituting this value:
\[ T_{5} = 330 \cdot \frac{128}{81} = \frac{42240}{81} \]

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