Find the 4th term from the end in the expansion of `((x)/(2)-(4)/(x))^(15)`

Explanation

To find the 4th term from the end in the expansion of \(\left(\frac{x}{2} - \frac{4}{x}\right)^{15}\), we can follow these steps:

### Step-by-Step Solution

1. **Identify the General Term**:
The general term \(T_r\) in the expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r-1} a^{n - (r - 1)} b^{r - 1} \]
Here, \(a = \frac{x}{2}\), \(b = -\frac{4}{x}\), and \(n = 15\).

2. **Determine the Term from the End**:
To find the 4th term from the end, we can use the relationship:
\[ T_{r} = T_{n - r + 1} \]
Therefore, the 4th term from the end corresponds to the \(n - 4 + 1 = 15 - 4 + 1 = 12\)th term from the beginning.

3. **Calculate the 12th Term**:
Using the formula for the general term:
\[ T_{12} = \binom{15}{12 - 1} \left(\frac{x}{2}\right)^{15 - (12 - 1)} \left(-\frac{4}{x}\right)^{12 - 1} \]
Simplifying this gives:
\[ T_{12} = \binom{15}{11} \left(\frac{x}{2}\right)^{4} \left(-\frac{4}{x}\right)^{11} \]

4. **Calculate the Binomial Coefficient**:
\[ \binom{15}{11} = \binom{15}{4} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365 \]

5. **Substitute Values**:
Now substituting the values:
\[ T_{12} = 1365 \left(\frac{x^4}{2^4}\right) \left(-\frac{4^{11}}{x^{11}}\right) \]
This simplifies to:
\[ T_{12} = 1365 \cdot \frac{x^4}{16} \cdot \left(-\frac{4^{11}}{x^{11}}\right) \]

6. **Combine Terms**:
\[ T_{12} = 1365 \cdot \left(-\frac{4^{11}}{16}\right) \cdot \frac{x^4}{x^{11}} = 1365 \cdot \left(-\frac{4^{11}}{16}\right) \cdot \frac{1}{x^7} \]

7. **Calculate Powers of 4**:
\[ -\frac{4^{11}}{16} = -\frac{4^{11}}{4^2} = -4^{9} = -262144 \]

8. **Final Expression**:
\[ T_{12} = 1365 \cdot (-262144) \cdot \frac{1}{x^7} \]
\[ T_{12} = -358318080 \cdot \frac{1}{x^7} \]

### Final Answer:
The 4th term from the end in the expansion of \(\left(\frac{x}{2} - \frac{4}{x}\right)^{15}\) is:
\[ -\frac{358318080}{x^7} \]