Find the numerically greatest terms in the expansion of
`(3x-4y)^(14)` when x = 8,y = 3

Answer

To find the numerically greatest terms in the expansion of \((3x - 4y)^{14}\) when \(x = 8\) and \(y = 3\), we can follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \(T_k\) in the expansion of \((a + b)^n\) is given by:
\[ T_k = \binom{n}{k} a^{n-k} b^k \]
In our case, \(a = 3x\), \(b = -4y\), and \(n = 14\). Therefore, the general term is:
\[ T_k = \binom{14}{k} (3x)^{14-k} (-4y)^k \]

### Step 2: Substitute the values of \(x\) and \(y\)
Substituting \(x = 8\) and \(y = 3\):
\[ T_k = \binom{14}{k} (3 \cdot 8)^{14-k} (-4 \cdot 3)^k \]
This simplifies to:
\[ T_k = \binom{14}{k} (24)^{14-k} (-12)^k \]

### Step 3: Find the ratio of consecutive terms
To find the numerically greatest term, we can use the ratio of consecutive terms:
\[ \frac{T_{k+1}}{T_k} = \frac{\binom{14}{k+1} (24)^{13-k} (-12)^{k+1}}{\binom{14}{k} (24)^{14-k} (-12)^k} \]
This simplifies to:
\[ \frac{T_{k+1}}{T_k} = \frac{14-k}{k+1} \cdot \frac{-12}{24} = \frac{14-k}{k+1} \cdot \left(-\frac{1}{2}\right) \]

### Step 4: Set the ratio to find the maximum
To find the greatest term, we set the ratio equal to \(-1\):
\[ \frac{14-k}{k+1} \cdot \left(-\frac{1}{2}\right) = -1 \]
This leads to:
\[ \frac{14-k}{k+1} = 2 \]
Cross-multiplying gives:
\[ 14 - k = 2(k + 1) \]
Expanding and simplifying:
\[ 14 - k = 2k + 2 \implies 14 - 2 = 3k \implies 12 = 3k \implies k = 4 \]

### Step 5: Identify the terms
The numerically greatest terms correspond to \(k = 4\) and \(k = 5\) (since \(k\) can be an integer and we look at \(k\) and \(k+1\)).

### Step 6: Write down the terms
Thus, the numerically greatest terms are:
- The 5th term (\(T_4\)):
\[ T_4 = \binom{14}{4} (24)^{10} (-12)^4 \]
- The 6th term (\(T_5\)):
\[ T_5 = \binom{14}{5} (24)^{9} (-12)^5 \]

### Conclusion
The numerically greatest terms in the expansion of \((3x - 4y)^{14}\) when \(x = 8\) and \(y = 3\) are the 5th and 6th terms.

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