Find the fifth expansion of `(a/3-3b)^(7)`

Answer

To find the fifth term in the expansion of \((\frac{a}{3} - 3b)^{7}\), we will use the Binomial Theorem. According to the Binomial Theorem, the general term \(T_{r+1}\) in the expansion of \((x + y)^n\) is given by:

\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]

In our case, we can identify:
- \(n = 7\)
- \(x = \frac{a}{3}\)
- \(y = -3b\)

### Step 1: Write the general term
The general term \(T_{r+1}\) for our expansion is:

\[ T_{r+1} = \binom{7}{r} \left(\frac{a}{3}\right)^{7-r} (-3b)^r \]

### Step 2: Find the fifth term
To find the fifth term, we need to calculate \(T_5\), which corresponds to \(r = 4\) (since \(T_{r+1}\) corresponds to \(r\)):

\[ T_5 = \binom{7}{4} \left(\frac{a}{3}\right)^{7-4} (-3b)^4 \]

### Step 3: Calculate the components
1. Calculate \(\binom{7}{4}\):

\[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

2. Calculate \(\left(\frac{a}{3}\right)^{3}\):

\[ \left(\frac{a}{3}\right)^{3} = \frac{a^3}{27} \]

3. Calculate \((-3b)^{4}\):

\[ (-3b)^{4} = 81b^4 \quad (\text{since } (-3)^4 = 81) \]

### Step 4: Combine the components
Now substitute these values back into the expression for \(T_5\):

\[ T_5 = 35 \cdot \frac{a^3}{27} \cdot 81b^4 \]

### Step 5: Simplify
Now, simplify \(T_5\):

\[ T_5 = 35 \cdot \frac{81}{27} a^3 b^4 \]

Calculating \(\frac{81}{27} = 3\):

\[ T_5 = 35 \cdot 3 a^3 b^4 = 105 a^3 b^4 \]

### Final Answer
Thus, the fifth term in the expansion of \((\frac{a}{3} - 3b)^{7}\) is:

\[ \boxed{105 a^3 b^4} \]