Find the numerically Greatest Term In the expansion of `(3-5x)^15` when x=1/5

Answer

To find the numerically greatest term in the expansion of \((3 - 5x)^{15}\) when \(x = \frac{1}{5}\), we can follow these steps:

### Step 1: Identify the General Term
The general term \(T_r\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = 3\), \(b = -5x\), and \(n = 15\). Thus, the general term becomes:
\[ T_r = \binom{15}{r} (3)^{15-r} (-5x)^r \]

### Step 2: Substitute \(x = \frac{1}{5}\)
Substituting \(x = \frac{1}{5}\) into the general term:
\[ T_r = \binom{15}{r} (3)^{15-r} (-5 \cdot \frac{1}{5})^r = \binom{15}{r} (3)^{15-r} (-1)^r \]

### Step 3: Simplify the General Term
This simplifies to:
\[ T_r = \binom{15}{r} (3)^{15-r} (-1)^r \]

### Step 4: Find the Ratio of Consecutive Terms
To find the greatest term, we can compare the ratio of consecutive terms \(T_r\) and \(T_{r+1}\):
\[ \frac{T_{r+1}}{T_r} = \frac{\binom{15}{r+1} (3)^{15-(r+1)} (-1)^{r+1}}{\binom{15}{r} (3)^{15-r} (-1)^r} \]
This simplifies to:
\[ \frac{T_{r+1}}{T_r} = \frac{\binom{15}{r+1}}{\binom{15}{r}} \cdot \frac{(3)^{14-r}}{(3)^{15-r}} \cdot (-1) = \frac{15 - r}{r + 1} \cdot \frac{1}{3} \cdot (-1) \]

### Step 5: Set the Ratio Greater Than or Equal to 1
To find the maximum term, we set the absolute value of the ratio greater than or equal to 1:
\[ \left|\frac{15 - r}{3(r + 1)}\right| \geq 1 \]
This leads to two inequalities:
1. \(15 - r \geq 3(r + 1)\)
2. \(15 - r \leq -3(r + 1)\)

### Step 6: Solve the Inequalities
1. From \(15 - r \geq 3(r + 1)\):
\[ 15 - r \geq 3r + 3 \implies 12 \geq 4r \implies r \leq 3 \]

2. From \(15 - r \leq -3(r + 1)\):
\[ 15 - r \leq -3r - 3 \implies 18 \leq 2r \implies r \geq 9 \]

### Step 7: Determine the Greatest Term
Since \(r\) must be an integer, the possible values for \(r\) are \(3\) and \(9\). We check the terms \(T_3\) and \(T_9\) to find the greatest term.

### Step 8: Calculate \(T_4\) (since \(T_{r+1}\) is the greatest term)
We find \(T_4\):
\[ T_4 = \binom{15}{4} (3)^{11} (-1)^4 = \binom{15}{4} (3)^{11} \]

### Step 9: Calculate the Value
Now we calculate:
\[ \binom{15}{4} = \frac{15!}{4!(15-4)!} = 1365 \]
So,
\[ T_4 = 1365 \cdot 3^{11} \]

### Final Answer
Thus, the numerically greatest term in the expansion of \((3 - 5x)^{15}\) when \(x = \frac{1}{5}\) is:
\[ T_4 = 1365 \cdot 3^{11} \]