Expand using Binomial Theorem `(1+x/2-2/x)^4,x!=0.`

Explanation

To expand the expression \((1 + \frac{x}{2} - \frac{2}{x})^4\) using the Binomial Theorem, we will follow these steps:

### Step 1: Identify the terms
We can rewrite the expression as:
\[ (1 + \frac{x}{2} - \frac{2}{x})^4 = (1 + \frac{x}{2} + (-\frac{2}{x}))^4 \]
Let \(a = 1\), \(b = \frac{x}{2}\), and \(c = -\frac{2}{x}\).

### Step 2: Apply the Binomial Theorem
According to the Binomial Theorem, the expansion of \((a + b + c)^n\) can be expressed as:
\[ \sum_{i+j+k=n} \frac{n!}{i!j!k!} a^i b^j c^k \]
where \(i + j + k = n\).

For our case, \(n = 4\), so we will consider all combinations of \(i\), \(j\), and \(k\) such that \(i + j + k = 4\).

### Step 3: Calculate the coefficients
We will calculate the contributions for each \(i\), \(j\), and \(k\):

1. **For \(i = 4, j = 0, k = 0\)**:
\[ \frac{4!}{4!0!0!} (1)^4 \left(\frac{x}{2}\right)^0 \left(-\frac{2}{x}\right)^0 = 1 \]

2. **For \(i = 3, j = 1, k = 0\)**:
\[ \frac{4!}{3!1!0!} (1)^3 \left(\frac{x}{2}\right)^1 \left(-\frac{2}{x}\right)^0 = 4 \cdot \frac{x}{2} = 2x \]

3. **For \(i = 2, j = 2, k = 0\)**:
\[ \frac{4!}{2!2!0!} (1)^2 \left(\frac{x}{2}\right)^2 \left(-\frac{2}{x}\right)^0 = 6 \cdot \frac{x^2}{4} = \frac{3x^2}{2} \]

4. **For \(i = 2, j = 1, k = 1\)**:
\[ \frac{4!}{2!1!1!} (1)^2 \left(\frac{x}{2}\right)^1 \left(-\frac{2}{x}\right)^1 = 12 \cdot \frac{x}{2} \cdot -\frac{2}{x} = -12 \]

5. **For \(i = 1, j = 3, k = 0\)**:
\[ \frac{4!}{1!3!0!} (1)^1 \left(\frac{x}{2}\right)^3 \left(-\frac{2}{x}\right)^0 = 4 \cdot \frac{x^3}{8} = \frac{x^3}{2} \]

6. **For \(i = 1, j = 2, k = 1\)**:
\[ \frac{4!}{1!2!1!} (1)^1 \left(\frac{x}{2}\right)^2 \left(-\frac{2}{x}\right)^1 = 12 \cdot \frac{x^2}{4} \cdot -\frac{2}{x} = -6x \]

7. **For \(i = 0, j = 4, k = 0\)**:
\[ \frac{4!}{0!4!0!} (1)^0 \left(\frac{x}{2}\right)^4 \left(-\frac{2}{x}\right)^0 = \frac{x^4}{16} \]

8. **For \(i = 0, j = 3, k = 1\)**:
\[ \frac{4!}{0!3!1!} (1)^0 \left(\frac{x}{2}\right)^3 \left(-\frac{2}{x}\right)^1 = -\frac{8}{x} \]

9. **For \(i = 0, j = 2, k = 2\)**:
\[ \frac{4!}{0!2!2!} (1)^0 \left(\frac{x}{2}\right)^2 \left(-\frac{2}{x}\right)^2 = 6 \cdot \frac{x^2}{4} \cdot \frac{4}{x^2} = 6 \]

10. **For \(i = 0, j = 1, k = 3\)**:
\[ \frac{4!}{0!1!3!} (1)^0 \left(\frac{x}{2}\right)^1 \left(-\frac{2}{x}\right)^3 = -\frac{8}{x^2} \]

11. **For \(i = 0, j = 0, k = 4\)**:
\[ \frac{4!}{0!0!4!} (1)^0 \left(\frac{x}{2}\right)^0 \left(-\frac{2}{x}\right)^4 = \frac{16}{x^4} \]

### Step 4: Combine all terms
Now, we can combine all the terms we calculated:
\[ 1 + 2x + \frac{3x^2}{2} - 12 + \frac{x^3}{2} - 6x + \frac{x^4}{16} - \frac{8}{x} + 6 - \frac{8}{x^2} + \frac{16}{x^4} \]

### Step 5: Simplify the expression
Combining like terms:
- Constant terms: \(1 - 12 + 6 = -5\)
- Coefficient of \(x\): \(2x - 6x = -4x\)
- Coefficient of \(x^2\): \(\frac{3x^2}{2} + 6 = \frac{3x^2 + 12}{2} = \frac{3x^2 + 12}{2}\)
- Coefficient of \(x^3\): \(\frac{x^3}{2}\)
- Coefficient of \(x^4\): \(\frac{x^4}{16}\)
- Coefficient of \(\frac{1}{x}\): \(-\frac{8}{x}\)
- Coefficient of \(\frac{1}{x^2}\): \(-\frac{8}{x^2}\)
- Coefficient of \(\frac{1}{x^4}\): \(\frac{16}{x^4}\)

Thus, the final expanded form is:
\[ \frac{x^4}{16} + \frac{x^3}{2} + \frac{3x^2 + 12}{2} - 4x - 5 - \frac{8}{x} - \frac{8}{x^2} + \frac{16}{x^4} \]