Find the numerically greatest term (s) in the expansion of
`(4+3x)^(15)" when "x=(7)/(2)`

Answer

To find the numerically greatest term in the expansion of \((4 + 3x)^{15}\) when \(x = \frac{7}{2}\), we can follow these steps:

### Step 1: Substitute the value of \(x\)
We start by substituting \(x = \frac{7}{2}\) into the expression:
\[ (4 + 3x)^{15} = (4 + 3 \cdot \frac{7}{2})^{15} = (4 + \frac{21}{2})^{15} = \left(\frac{8}{2} + \frac{21}{2}\right)^{15} = \left(\frac{29}{2}\right)^{15} \]

### Step 2: Rewrite the expression
Now, we can express the expansion as:
\[ \left(\frac{29}{2}\right)^{15} = \frac{29^{15}}{2^{15}} \]

### Step 3: Use the Binomial Theorem
Using the Binomial Theorem, we can expand \((4 + 3x)^{15}\):
\[ (4 + 3x)^{15} = \sum_{k=0}^{15} \binom{15}{k} (4)^{15-k} (3x)^{k} \]

### Step 4: Identify the term
The \(k\)-th term in the expansion is given by:
\[ T_k = \binom{15}{k} (4)^{15-k} (3x)^{k} \]
Substituting \(x = \frac{7}{2}\):
\[ T_k = \binom{15}{k} (4)^{15-k} (3 \cdot \frac{7}{2})^{k} = \binom{15}{k} (4)^{15-k} \left(\frac{21}{2}\right)^{k} \]

### Step 5: Find the ratio of consecutive terms
To find the numerically greatest term, we can find the ratio of consecutive terms \(T_k\) and \(T_{k+1}\):
\[ \frac{T_{k+1}}{T_k} = \frac{\binom{15}{k+1} (4)^{15-(k+1)} \left(\frac{21}{2}\right)^{k+1}}{\binom{15}{k} (4)^{15-k} \left(\frac{21}{2}\right)^{k}} = \frac{15-k}{k+1} \cdot \frac{(21/2)}{4} \]

### Step 6: Set the ratio to 1
To find the maximum term, we set the ratio equal to 1:
\[ \frac{15-k}{k+1} \cdot \frac{21}{8} = 1 \]
This leads to:
\[ 15 - k = \frac{8(k+1)}{21} \]
Cross-multiplying gives:
\[ 21(15 - k) = 8(k + 1) \]
\[ 315 - 21k = 8k + 8 \]
\[ 315 - 8 = 21k + 8k \]
\[ 307 = 29k \implies k = \frac{307}{29} \approx 10.586 \]

### Step 7: Find the integer values of \(k\)
Since \(k\) must be an integer, we check \(k = 10\) and \(k = 11\).

### Step 8: Calculate the terms
Calculate \(T_{10}\) and \(T_{11}\):
- For \(k = 10\):
\[ T_{10} = \binom{15}{10} (4)^{5} \left(\frac{21}{2}\right)^{10} \]
- For \(k = 11\):
\[ T_{11} = \binom{15}{11} (4)^{4} \left(\frac{21}{2}\right)^{11} \]

### Step 9: Compare the terms
Compare \(T_{10}\) and \(T_{11}\) to find which is greater.

### Final Result
The numerically greatest term in the expansion of \((4 + 3x)^{15}\) when \(x = \frac{7}{2}\) is either \(T_{10}\) or \(T_{11}\), depending on which is larger.