The numerically greatest term in the expansion `(2x - 3y)^12` when x = 1 and y = 5/2 is the

Options

A. 11th term
B. 10th term
C. 9th term
D. 6th term

Answer

To find the numerically greatest term in the expansion of \((2x - 3y)^{12}\) when \(x = 1\) and \(y = \frac{5}{2}\), we can follow these steps:

### Step 1: Identify the General Term
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, \(a = 2x\), \(b = -3y\), and \(n = 12\). Thus, the general term becomes:
\[ T_k = \binom{12}{k} (2x)^{12-k} (-3y)^k \]

### Step 2: Substitute the Values of \(x\) and \(y\)
Substituting \(x = 1\) and \(y = \frac{5}{2}\):
\[ T_k = \binom{12}{k} (2 \cdot 1)^{12-k} (-3 \cdot \frac{5}{2})^k \]
This simplifies to:
\[ T_k = \binom{12}{k} 2^{12-k} (-\frac{15}{2})^k \]
\[ = \binom{12}{k} 2^{12-k} \left(-\frac{15}{2}\right)^k \]
\[ = \binom{12}{k} 2^{12-k} \cdot (-1)^k \cdot \frac{15^k}{2^k} \]
\[ = \binom{12}{k} (-1)^k \cdot 15^k \cdot 2^{12-k} \cdot 2^{-k} \]
\[ = \binom{12}{k} (-1)^k \cdot 15^k \cdot 2^{12-2k} \]

### Step 3: Find the Absolute Value of the General Term
To find the numerically greatest term, we consider the absolute value:
\[ |T_k| = \binom{12}{k} \cdot 15^k \cdot 2^{12-2k} \]

### Step 4: Determine the Value of \(k\) for Maximum Term
To find the maximum term, we can use the ratio of consecutive terms:
\[ \frac{|T_{k+1}|}{|T_k|} = \frac{\binom{12}{k+1} \cdot 15^{k+1} \cdot 2^{12-2(k+1)}}{\binom{12}{k} \cdot 15^k \cdot 2^{12-2k}} = \frac{\binom{12}{k+1}}{\binom{12}{k}} \cdot \frac{15}{2^2} \]
\[ = \frac{12-k}{k+1} \cdot \frac{15}{4} \]

Setting the ratio equal to 1 to find the maximum:
\[ \frac{12-k}{k+1} \cdot \frac{15}{4} = 1 \]
\[ 12-k = \frac{4(k+1)}{15} \]
Cross-multiplying gives:
\[ 15(12-k) = 4(k+1) \]
\[ 180 - 15k = 4k + 4 \]
\[ 180 - 4 = 15k + 4k \]
\[ 176 = 19k \implies k = \frac{176}{19} \approx 9.26 \]

### Step 5: Determine the Integer Value of \(k\)
Since \(k\) must be an integer, we check \(k = 9\) and \(k = 10\).

### Step 6: Calculate \(T_9\) and \(T_{10}\)
We can calculate both terms to see which is greater:
- For \(k = 9\):
\[ |T_9| = \binom{12}{9} \cdot 15^9 \cdot 2^{12-18} \]
- For \(k = 10\):
\[ |T_{10}| = \binom{12}{10} \cdot 15^{10} \cdot 2^{12-20} \]

### Step 7: Compare the Two Terms
After calculating, we find that the term with the highest absolute value is \(T_{10}\).

### Conclusion
The numerically greatest term in the expansion \((2x - 3y)^{12}\) when \(x = 1\) and \(y = \frac{5}{2}\) is:
\[ \text{The 11th term } (T_{10}). \]