The algebraically second largest term in the expansion of `(3-2x)^(15)` at `x=(4)/(3)`.

Options

A. `5`
B. `7`
C. `9`
D. `11`

Explanation

To find the algebraically second largest term in the expansion of \((3 - 2x)^{15}\) at \(x = \frac{4}{3}\), we will follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \(T_k\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_k = \binom{n}{k} a^{n-k} b^k \]
For our case, \(a = 3\), \(b = -2x\), and \(n = 15\). Thus, the general term becomes:
\[ T_k = \binom{15}{k} (3)^{15-k} (-2x)^k \]

### Step 2: Substitute \(x = \frac{4}{3}\)
Substituting \(x = \frac{4}{3}\) into the general term:
\[ T_k = \binom{15}{k} (3)^{15-k} (-2 \cdot \frac{4}{3})^k \]
This simplifies to:
\[ T_k = \binom{15}{k} (3)^{15-k} \left(-\frac{8}{3}\right)^k \]
\[ T_k = \binom{15}{k} (3)^{15-k} \cdot \frac{(-8)^k}{3^k} \]
\[ T_k = \binom{15}{k} (-8)^k (3)^{15-k} \cdot 3^{-k} \]
\[ T_k = \binom{15}{k} (-8)^k \cdot 3^{15-2k} \]

### Step 3: Determine the largest term
To find the largest term, we need to analyze the ratio of consecutive terms:
\[ \frac{T_{k+1}}{T_k} = \frac{\binom{15}{k+1} (-8)^{k+1} 3^{15-2(k+1)}}{\binom{15}{k} (-8)^k 3^{15-2k}} \]
This simplifies to:
\[ \frac{T_{k+1}}{T_k} = \frac{15-k}{k+1} \cdot \frac{-8}{3^2} = \frac{15-k}{k+1} \cdot \frac{-8}{9} \]
Setting this ratio greater than 1 to find the maximum term:
\[ \frac{15-k}{k+1} \cdot \frac{-8}{9} > 1 \]
This leads to:
\[ \frac{15-k}{k+1} > -\frac{9}{8} \]

### Step 4: Solve for \(k\)
This inequality can be solved for \(k\) to find the range of \(k\) values that yield the largest terms. However, we can also find the approximate value of \(k\) by trial or by estimating:
\[ k \approx \frac{n}{2} = \frac{15}{2} \approx 7.5 \]
Thus, we check \(k = 7\) and \(k = 8\).

### Step 5: Calculate \(T_7\) and \(T_8\)
Calculate \(T_7\):
\[ T_7 = \binom{15}{7} (-8)^7 \cdot 3^{15-14} = \binom{15}{7} (-8)^7 \cdot 3^1 \]
Calculate \(T_8\):
\[ T_8 = \binom{15}{8} (-8)^8 \cdot 3^{15-16} = \binom{15}{8} (-8)^8 \cdot 3^{-1} \]

### Step 6: Compare \(T_7\) and \(T_8\)
Since \(T_8\) is a positive term and \(T_7\) is negative, we can conclude that \(T_8\) is the largest term and \(T_7\) is the second largest term.

### Conclusion
The algebraically second largest term in the expansion of \((3 - 2x)^{15}\) at \(x = \frac{4}{3}\) is \(T_7\).