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

Options

A. `(189)/(16)x^(14)`
B. `(21)/(8)x^(13)`
C. Both a & b
D. none of these

Answer

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

### Step 1: Identify the value of \(n\)
In the expression \(\left(a + b\right)^n\), we have:
- \(a = \frac{3x^2}{2}\)
- \(b = \frac{x}{3}\)
- \(n = 3\)

### Step 2: Determine the number of terms in the expansion
The number of terms in the expansion of \((a + b)^n\) is \(n + 1\). Therefore, for \(n = 3\):
\[ \text{Number of terms} = 3 + 1 = 4 \]

### Step 3: Find the middle term(s)
Since there are 4 terms, the middle terms will be the 2nd and 3rd terms (as \(n\) is even). We will denote the \(k\)-th term in the expansion as:
\[ T_k = \binom{n}{k} a^{n-k} b^k \]
where \(k\) starts from 0.

### Step 4: Calculate the 2nd term (\(T_2\))
For \(T_2\) (where \(k = 1\)):
\[ T_2 = \binom{3}{1} \left(\frac{3x^2}{2}\right)^{3-1} \left(\frac{x}{3}\right)^1 \]
Calculating this:
\[ T_2 = \binom{3}{1} \left(\frac{3x^2}{2}\right)^{2} \left(\frac{x}{3}\right) \]
\[ = 3 \cdot \left(\frac{9x^4}{4}\right) \cdot \left(\frac{x}{3}\right) \]
\[ = 3 \cdot \frac{9x^4}{4} \cdot \frac{x}{3} \]
\[ = \frac{9x^5}{4} \]

### Step 5: Calculate the 3rd term (\(T_3\))
For \(T_3\) (where \(k = 2\)):
\[ T_3 = \binom{3}{2} \left(\frac{3x^2}{2}\right)^{3-2} \left(\frac{x}{3}\right)^2 \]
Calculating this:
\[ T_3 = \binom{3}{2} \left(\frac{3x^2}{2}\right)^{1} \left(\frac{x}{3}\right)^{2} \]
\[ = 3 \cdot \left(\frac{3x^2}{2}\right) \cdot \left(\frac{x^2}{9}\right) \]
\[ = 3 \cdot \frac{3x^2}{2} \cdot \frac{x^2}{9} \]
\[ = 3 \cdot \frac{3x^4}{18} \]
\[ = \frac{3x^4}{6} = \frac{x^4}{2} \]

### Final Result
The middle terms in the expansion are:
- \(T_2 = \frac{9x^5}{4}\)
- \(T_3 = \frac{x^4}{2}\)