Find the numerically greatest term in the expansion of `(3-2x)^9` when x=1

Answer

To find the numerically greatest term in the expansion of \((3 - 2x)^9\) when \(x = 1\), we will follow these steps:

### Step 1: Substitute \(x = 1\)
First, we substitute \(x = 1\) into the expression:
\[ (3 - 2x)^9 = (3 - 2 \cdot 1)^9 = (3 - 2)^9 = 1^9 = 1 \]
However, we need to find the greatest term in the binomial expansion.

### Step 2: Write the Binomial Expansion
The binomial expansion of \((a - b)^n\) is given by:
\[ \sum_{r=0}^{n} \binom{n}{r} a^{n-r} (-b)^r \]
In our case, \(a = 3\), \(b = 2\), and \(n = 9\). Thus, the expansion is:
\[ (3 - 2)^9 = \sum_{r=0}^{9} \binom{9}{r} 3^{9-r} (-2)^r \]

### Step 3: Find the General Term
The general term \(T_{r+1}\) in the expansion can be expressed as:
\[ T_{r+1} = \binom{9}{r} 3^{9-r} (-2)^r \]
To find the numerically greatest term, we will consider the absolute value:
\[ |T_{r+1}| = \binom{9}{r} 3^{9-r} 2^r \]

### Step 4: Find the Ratio of Consecutive Terms
To find the maximum term, we can compare consecutive terms \(T_r\) and \(T_{r+1}\). We will find the ratio:
\[ \frac{T_{r+1}}{T_r} = \frac{\binom{9}{r+1} 3^{9-(r+1)} (-2)^{r+1}}{\binom{9}{r} 3^{9-r} (-2)^r} \]
This simplifies to:
\[ \frac{T_{r+1}}{T_r} = \frac{9-r}{r+1} \cdot \frac{-2}{3} \]

### Step 5: Set the Ratio to 1
To find the maximum term, we set the absolute value of the ratio to 1:
\[ \left|\frac{9-r}{r+1} \cdot \frac{-2}{3}\right| = 1 \]
This leads to:
\[ \frac{9-r}{r+1} \cdot \frac{2}{3} = 1 \]
Cross-multiplying gives:
\[ 2(9 - r) = 3(r + 1) \]
Expanding and rearranging:
\[ 18 - 2r = 3r + 3 \implies 18 - 3 = 5r \implies 15 = 5r \implies r = 3 \]

### Step 6: Find the Greatest Term
The greatest term occurs when \(r = 4\) (the next integer). Thus, we calculate \(T_4\):
\[ T_4 = \binom{9}{4} 3^{9-4} (-2)^4 \]
Calculating each part:
\[ \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \]
\[ 3^{5} = 243 \]
\[ (-2)^4 = 16 \]
Now, substituting these values:
\[ T_4 = 126 \times 243 \times 16 \]

### Step 7: Calculate the Final Value
Calculating \(T_4\):
\[ T_4 = 126 \times 243 \times 16 = 489888 \]

### Conclusion
The numerically greatest term in the expansion of \((3 - 2x)^9\) when \(x = 1\) is:
\[ \boxed{489888} \]