Expand the binomial `((2)/(x) +x)^(10)` up to four terms

Answer

To expand the binomial expression \(\left(\frac{2}{x} + x\right)^{10}\) up to four terms, we will use the Binomial Theorem. The Binomial Theorem states that:

\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]

In our case, we can identify:
- \(x = \frac{2}{x}\)
- \(y = x\)
- \(n = 10\)

Now, we will calculate the first four terms of the expansion.

### Step 1: Calculate the first term
For \(k = 0\):
\[ \text{Term}_0 = \binom{10}{0} \left(\frac{2}{x}\right)^{10} (x)^0 = 1 \cdot \left(\frac{2}{x}\right)^{10} \cdot 1 = \frac{2^{10}}{x^{10}} = \frac{1024}{x^{10}} \]

### Step 2: Calculate the second term
For \(k = 1\):
\[ \text{Term}_1 = \binom{10}{1} \left(\frac{2}{x}\right)^{9} (x)^1 = 10 \cdot \left(\frac{2}{x}\right)^{9} \cdot x = 10 \cdot \frac{2^9}{x^9} \cdot x = 10 \cdot \frac{512}{x^9} \cdot x = \frac{5120}{x^8} \]

### Step 3: Calculate the third term
For \(k = 2\):
\[ \text{Term}_2 = \binom{10}{2} \left(\frac{2}{x}\right)^{8} (x)^2 = 45 \cdot \left(\frac{2}{x}\right)^{8} \cdot x^2 = 45 \cdot \frac{2^8}{x^8} \cdot x^2 = 45 \cdot \frac{256}{x^8} \cdot x^2 = \frac{11520}{x^6} \]

### Step 4: Calculate the fourth term
For \(k = 3\):
\[ \text{Term}_3 = \binom{10}{3} \left(\frac{2}{x}\right)^{7} (x)^3 = 120 \cdot \left(\frac{2}{x}\right)^{7} \cdot x^3 = 120 \cdot \frac{2^7}{x^7} \cdot x^3 = 120 \cdot \frac{128}{x^7} \cdot x^3 = \frac{15360}{x^4} \]

### Final Expansion
Now, we can combine all four terms we calculated:
\[ \left(\frac{2}{x} + x\right)^{10} \approx \frac{1024}{x^{10}} + \frac{5120}{x^8} + \frac{11520}{x^6} + \frac{15360}{x^4} \]

### Summary of the Expansion
Thus, the expansion of \(\left(\frac{2}{x} + x\right)^{10}\) up to four terms is:
\[ \frac{1024}{x^{10}} + \frac{5120}{x^8} + \frac{11520}{x^6} + \frac{15360}{x^4} \]