Find the middle term in the expansion of `(2x + 3/x)^20`

Answer

To find the middle term in the expansion of \((2x + \frac{3}{x})^{20}\), we can follow these steps:

### Step 1: Identify the total number of terms
In the expansion of \((A + B)^n\), the number of terms is given by \(n + 1\). Here, \(n = 20\), so the number of terms is:
\[ 20 + 1 = 21 \]

### Step 2: Determine the middle term
Since the total number of terms (21) is odd, the middle term will be the \(\frac{n}{2} + 1\)th term. Thus, the middle term is the \(11\)th term.

### Step 3: Use the Binomial Theorem
The \(r\)th term in the expansion of \((A + B)^n\) is given by:
\[ T_{r} = \binom{n}{r-1} A^{n-(r-1)} B^{r-1} \]
For the 11th term (\(r = 11\)):
\[ T_{11} = \binom{20}{10} (2x)^{20-10} \left(\frac{3}{x}\right)^{10} \]

### Step 4: Substitute values into the formula
Substituting the values into the formula:
\[ T_{11} = \binom{20}{10} (2x)^{10} \left(\frac{3}{x}\right)^{10} \]

### Step 5: Simplify the expression
Now, we simplify \(T_{11}\):
\[ T_{11} = \binom{20}{10} (2^{10} x^{10}) \left(\frac{3^{10}}{x^{10}}\right) \]
The \(x^{10}\) terms cancel out:
\[ T_{11} = \binom{20}{10} \cdot 2^{10} \cdot 3^{10} \]

### Step 6: Calculate \(\binom{20}{10}\)
Using the formula for combinations:
\[ \binom{20}{10} = \frac{20!}{10! \cdot 10!} \]

### Step 7: Calculate \(2^{10}\) and \(3^{10}\)
Calculating the powers:
\[ 2^{10} = 1024 \quad \text{and} \quad 3^{10} = 59049 \]

### Step 8: Combine all parts
Now, we can calculate \(T_{11}\):
\[ T_{11} = \frac{20!}{10! \cdot 10!} \cdot 1024 \cdot 59049 \]

### Step 9: Final calculation
Calculating \(\binom{20}{10}\):
\[ \binom{20}{10} = 184756 \]
Thus,
\[ T_{11} = 184756 \cdot 1024 \cdot 59049 \]

### Step 10: Final Result
Calculating the final value:
\[ T_{11} = 184756 \cdot 1024 \cdot 59049 = 184756 \cdot 60466176 = 11112006825558016 \]

### Conclusion
The middle term in the expansion of \((2x + \frac{3}{x})^{20}\) is:
\[ T_{11} = 11112006825558016 \]