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

Answer

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

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

where \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\).

### Step 1: Identify \(a\), \(b\), and \(n\)
In our case, we have:
- \(a = \frac{2}{x}\)
- \(b = x\)
- \(n = 10\)

### Step 2: Write the first four terms of the expansion
Using the Binomial Theorem, we will calculate the first four terms (for \(k = 0, 1, 2, 3\)):

1. **First term** (\(k = 0\)):
\[ T_0 = \binom{10}{0} \left(\frac{2}{x}\right)^{10} x^0 = 1 \cdot \left(\frac{2}{x}\right)^{10} \cdot 1 = \frac{2^{10}}{x^{10}} = \frac{1024}{x^{10}} \]

2. **Second term** (\(k = 1\)):
\[ T_1 = \binom{10}{1} \left(\frac{2}{x}\right)^{9} x^1 = 10 \cdot \left(\frac{2}{x}\right)^{9} \cdot x = 10 \cdot \frac{2^9}{x^9} \cdot x = 10 \cdot \frac{512}{x^9} \cdot x = \frac{5120}{x^8} \]

3. **Third term** (\(k = 2\)):
\[ T_2 = \binom{10}{2} \left(\frac{2}{x}\right)^{8} x^2 = 45 \cdot \left(\frac{2}{x}\right)^{8} \cdot x^2 = 45 \cdot \frac{2^8}{x^8} \cdot x^2 = 45 \cdot \frac{256}{x^8} \cdot x^2 = \frac{11520}{x^6} \]

4. **Fourth term** (\(k = 3\)):
\[ T_3 = \binom{10}{3} \left(\frac{2}{x}\right)^{7} x^3 = 120 \cdot \left(\frac{2}{x}\right)^{7} \cdot x^3 = 120 \cdot \frac{2^7}{x^7} \cdot x^3 = 120 \cdot \frac{128}{x^7} \cdot x^3 = \frac{15360}{x^4} \]

### Step 3: Combine the terms
Now, we can combine the first four terms we calculated:

\[ \left(\frac{2}{x} + x\right)^{10} \approx \frac{1024}{x^{10}} + \frac{5120}{x^8} + \frac{11520}{x^6} + \frac{15360}{x^4} \]

### Final Answer
Thus, the expansion of \(\left(\frac{2}{x} + x\right)^{10}\) up to four terms is:

\[ \frac{1024}{x^{10}} + \frac{5120}{x^8} + \frac{11520}{x^6} + \frac{15360}{x^4} \]