Write the middle term in the expansion of `((2x^2)/3+3/(2x^2))^(10)dot`

Options

A. `"^9C_5`
B. `"^10C_5`
C. `"^10C_6`
D. none of these

Explanation

To find the middle term in the expansion of \(\left(\frac{2x^2}{3} + \frac{3}{2x^2}\right)^{10}\), we will follow these steps:

### Step 1: Identify \(n\) and determine the middle term
The given expression is raised to the power of \(n = 10\). Since \(n\) is even, the middle term can be found using the formula:
\[ \text{Middle Term} = T\left(\frac{n}{2} + 1\right) \]
Here, \(\frac{n}{2} + 1 = \frac{10}{2} + 1 = 6\). Therefore, we need to find \(T_6\).

### Step 2: Write the general term
The general term \(T_k\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_k = \binom{n}{k-1} a^{n-(k-1)} b^{k-1} \]
In our case, \(a = \frac{2x^2}{3}\), \(b = \frac{3}{2x^2}\), and \(n = 10\).

### Step 3: Substitute values to find \(T_6\)
We substitute \(k = 6\) into the general term formula:
\[ T_6 = \binom{10}{6-1} \left(\frac{2x^2}{3}\right)^{10-5} \left(\frac{3}{2x^2}\right)^{5} \]
This simplifies to:
\[ T_6 = \binom{10}{5} \left(\frac{2x^2}{3}\right)^{5} \left(\frac{3}{2x^2}\right)^{5} \]

### Step 4: Simplify the expression
Now we simplify the expression:
\[ T_6 = \binom{10}{5} \left(\frac{2^5 (x^2)^5}{3^5}\right) \left(\frac{3^5}{(2x^2)^5}\right) \]
\[ = \binom{10}{5} \cdot \frac{2^5 \cdot 3^5 \cdot (x^{10})}{3^5 \cdot 2^5 \cdot (x^{10})} \]
Notice that \(2^5\) and \(3^5\) cancel out, and \(x^{10}\) also cancels out:
\[ T_6 = \binom{10}{5} \]

### Step 5: Calculate \(\binom{10}{5}\)
Now we compute \(\binom{10}{5}\):
\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]

### Conclusion
Thus, the middle term in the expansion is:
\[ \text{Middle Term} = 252 \]