The numerically greatest term in the expansion `(5x - 6y)^14` when x = 2/5, y = 1/2 is

Options

A. `""^14C_(6)2^(8) 3^6`
B. `""^14C_(7) 2^(6)3^8`
C. `""^14C_(6) 2^(6) 3^8`
D. `""^14C_(7)2^(8)3^6`

Answer

To find the numerically greatest term in the expansion of \((5x - 6y)^{14}\) when \(x = \frac{2}{5}\) and \(y = \frac{1}{2}\), we will follow these steps:

### Step 1: Identify the General Term
The general term \(T_{r+1}\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our case, \(a = 5x\) and \(b = -6y\), and \(n = 14\).

### Step 2: Substitute Values
Substituting \(a\) and \(b\) into the formula, we have:
\[ T_{r+1} = \binom{14}{r} (5x)^{14-r} (-6y)^r \]
Now substituting \(x = \frac{2}{5}\) and \(y = \frac{1}{2}\):
\[ T_{r+1} = \binom{14}{r} \left(5 \cdot \frac{2}{5}\right)^{14-r} \left(-6 \cdot \frac{1}{2}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{14}{r} (2)^{14-r} (-3)^r \]

### Step 3: Find the Absolute Value
To find the numerically greatest term, we need to consider the absolute value:
\[ |T_{r+1}| = \binom{14}{r} (2)^{14-r} (3)^r \]

### Step 4: Determine the Ratio of Consecutive Terms
To find the maximum term, we will analyze the ratio of consecutive terms:
\[ \frac{|T_{r+1}|}{|T_r|} = \frac{\binom{14}{r} (2)^{14-r} (3)^r}{\binom{14}{r-1} (2)^{14-(r-1)} (3)^{r-1}} \]
This simplifies to:
\[ \frac{|T_{r+1}|}{|T_r|} = \frac{14-r+1}{r} \cdot \frac{3}{2} \]
Setting this ratio to 1 to find the maximum:
\[ \frac{14 - r + 1}{r} \cdot \frac{3}{2} = 1 \]

### Step 5: Solve for \(r\)
Cross-multiplying gives:
\[ (15 - r) \cdot 3 = 2r \]
\[ 45 - 3r = 2r \]
\[ 45 = 5r \implies r = 9 \]

### Step 6: Find the Greatest Term
Now, we substitute \(r = 9\) back into the term formula:
\[ T_{10} = \binom{14}{9} (2)^{14-9} (-3)^9 \]
Calculating:
\[ T_{10} = \binom{14}{9} (2)^5 (-3)^9 \]
\[ = \binom{14}{9} \cdot 32 \cdot (-19683) \]

### Step 7: Calculate the Coefficient
The binomial coefficient \(\binom{14}{9} = \binom{14}{5} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002\).

### Step 8: Final Calculation
Thus, the term becomes:
\[ T_{10} = 2002 \cdot 32 \cdot (-19683) \]
Calculating this gives:
\[ T_{10} = 2002 \cdot 32 \cdot 19683 \]

### Final Answer
The numerically greatest term in the expansion \((5x - 6y)^{14}\) when \(x = \frac{2}{5}\) and \(y = \frac{1}{2}\) is:
\[ 2002 \cdot 32 \cdot 19683 \]