The value of the numerically greatest term in the expansion of `(4-3x)^(7)` when `x=(2)/(3)` is equal to

Options

A. 71680
B. 35840
C. 10752
D. 86016

Answer

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

### Step 1: Identify the General Term
The general term \(T_{r+1}\) in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our case, \(a = 4\), \(b = -3x\), and \(n = 7\). Therefore, the general term becomes:
\[ T_{r+1} = \binom{7}{r} (4)^{7-r} (-3x)^r \]

### Step 2: Substitute \(x\)
Substituting \(x = \frac{2}{3}\) into the term gives:
\[ T_{r+1} = \binom{7}{r} (4)^{7-r} \left(-3 \cdot \frac{2}{3}\right)^r = \binom{7}{r} (4)^{7-r} (-2)^r \]

### Step 3: Find the Condition for the Greatest Term
The term \(T_{r+1}\) is numerically greatest when:
\[ r = \left\lfloor \frac{n + 1}{2} \cdot \frac{|b|}{|a| + |b|} \right\rfloor \]
Here, \(n = 7\), \(|a| = 4\), and \(|b| = 3x = 2\). Thus:
\[ r = \left\lfloor \frac{7 + 1}{2} \cdot \frac{2}{4 + 2} \right\rfloor = \left\lfloor 4 \cdot \frac{2}{6} \right\rfloor = \left\lfloor \frac{8}{6} \right\rfloor = \left\lfloor \frac{4}{3} \right\rfloor = 1 \]

### Step 4: Calculate the Greatest Term
Now, we need to find \(T_{r+1}\) where \(r = 2\):
\[ T_{3} = \binom{7}{2} (4)^{7-2} (-2)^2 \]
Calculating each part:
- \(\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21\)
- \((4)^{5} = 1024\)
- \((-2)^{2} = 4\)

Putting it all together:
\[ T_{3} = 21 \cdot 1024 \cdot 4 = 21 \cdot 4096 \]
Calculating \(21 \cdot 4096\):
\[ 21 \cdot 4096 = 86016 \]

### Final Answer
The value of the numerically greatest term in the expansion of \((4 - 3x)^{7}\) when \(x = \frac{2}{3}\) is:
\[ \boxed{86016} \]