Find the middle term of `(3/2x-2/5y)^(10)`

Answer

To find the middle term of the expression \((\frac{3}{2}x - \frac{2}{5}y)^{10}\), we will follow these steps:

### Step 1: Identify the Values
In the expression \((a + b)^n\), we have:
- \(a = \frac{3}{2}x\)
- \(b = -\frac{2}{5}y\)
- \(n = 10\)

### Step 2: Determine the Middle Term
For a binomial expansion, if \(n\) is even, the middle term is given by the \(\left(\frac{n}{2} + 1\right)\)th term. Since \(n = 10\) (which is even), we calculate:
\[ \text{Middle term} = \left(\frac{10}{2} + 1\right) = 5 + 1 = 6 \]
Thus, we need to find the 6th term of the expansion.

### Step 3: Write the General Term
The general term \(T_r\) in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our case:
\[ T_{r+1} = \binom{10}{r} \left(\frac{3}{2}x\right)^{10-r} \left(-\frac{2}{5}y\right)^r \]

### Step 4: Substitute for the 6th Term
To find the 6th term, we set \(r = 5\) (since \(T_{r+1}\) corresponds to \(T_6\)):
\[ T_6 = \binom{10}{5} \left(\frac{3}{2}x\right)^{10-5} \left(-\frac{2}{5}y\right)^5 \]

### Step 5: Calculate Each Component
1. Calculate \(\binom{10}{5}\):
\[ \binom{10}{5} = \frac{10!}{5!5!} = 252 \]

2. Calculate \(\left(\frac{3}{2}x\right)^5\):
\[ \left(\frac{3}{2}x\right)^5 = \left(\frac{3^5}{2^5}\right)x^5 = \frac{243}{32}x^5 \]

3. Calculate \(\left(-\frac{2}{5}y\right)^5\):
\[ \left(-\frac{2}{5}y\right)^5 = -\frac{32}{3125}y^5 \]

### Step 6: Combine the Results
Now, substitute these values back into the expression for \(T_6\):
\[ T_6 = 252 \cdot \frac{243}{32}x^5 \cdot \left(-\frac{32}{3125}y^5\right) \]

### Step 7: Simplify
\[ T_6 = 252 \cdot \frac{243 \cdot (-32)}{32 \cdot 3125} x^5 y^5 \]
The \(32\) cancels out:
\[ T_6 = 252 \cdot \frac{243 \cdot (-1)}{3125} x^5 y^5 \]
Calculating:
\[ T_6 = -\frac{243 \cdot 252}{3125} x^5 y^5 \]

### Final Result
The middle term of \((\frac{3}{2}x - \frac{2}{5}y)^{10}\) is:
\[ -\frac{61236}{3125} x^5 y^5 \]