Expand (i) `(x+1/x)^(7)` , (ii) `(x^(2)+2/x)^(4)` using binomial theorem.

Answer

To expand the expressions using the Binomial Theorem, we will follow the general formula for binomial expansion, which states:

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

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

### Part (i): Expand \((x + \frac{1}{x})^7\)

1. **Identify \(a\), \(b\), and \(n\)**:
- Here, \(a = x\), \(b = \frac{1}{x}\), and \(n = 7\).

2. **Write the expansion**:
\[ (x + \frac{1}{x})^7 = \sum_{r=0}^{7} \binom{7}{r} x^{7-r} \left(\frac{1}{x}\right)^r \]

3. **Calculate each term**:
- For \(r = 0\):
\[ \binom{7}{0} x^{7-0} \left(\frac{1}{x}\right)^0 = 1 \cdot x^7 \cdot 1 = x^7 \]
- For \(r = 1\):
\[ \binom{7}{1} x^{7-1} \left(\frac{1}{x}\right)^1 = 7 \cdot x^6 \cdot \frac{1}{x} = 7x^5 \]
- For \(r = 2\):
\[ \binom{7}{2} x^{7-2} \left(\frac{1}{x}\right)^2 = 21 \cdot x^5 \cdot \frac{1}{x^2} = 21x^3 \]
- For \(r = 3\):
\[ \binom{7}{3} x^{7-3} \left(\frac{1}{x}\right)^3 = 35 \cdot x^4 \cdot \frac{1}{x^3} = 35x \]
- For \(r = 4\):
\[ \binom{7}{4} x^{7-4} \left(\frac{1}{x}\right)^4 = 35 \cdot x^3 \cdot \frac{1}{x^4} = \frac{35}{x} \]
- For \(r = 5\):
\[ \binom{7}{5} x^{7-5} \left(\frac{1}{x}\right)^5 = 21 \cdot x^2 \cdot \frac{1}{x^5} = \frac{21}{x^3} \]
- For \(r = 6\):
\[ \binom{7}{6} x^{7-6} \left(\frac{1}{x}\right)^6 = 7 \cdot x^1 \cdot \frac{1}{x^6} = \frac{7}{x^5} \]
- For \(r = 7\):
\[ \binom{7}{7} x^{7-7} \left(\frac{1}{x}\right)^7 = 1 \cdot 1 \cdot \frac{1}{x^7} = \frac{1}{x^7} \]

4. **Combine all terms**:
\[ (x + \frac{1}{x})^7 = x^7 + 7x^5 + 21x^3 + 35x + \frac{35}{x} + \frac{21}{x^3} + \frac{7}{x^5} + \frac{1}{x^7} \]

### Part (ii): Expand \((x^2 + \frac{2}{x})^4\)

1. **Identify \(a\), \(b\), and \(n\)**:
- Here, \(a = x^2\), \(b = \frac{2}{x}\), and \(n = 4\).

2. **Write the expansion**:
\[ (x^2 + \frac{2}{x})^4 = \sum_{r=0}^{4} \binom{4}{r} (x^2)^{4-r} \left(\frac{2}{x}\right)^r \]

3. **Calculate each term**:
- For \(r = 0\):
\[ \binom{4}{0} (x^2)^{4-0} \left(\frac{2}{x}\right)^0 = 1 \cdot x^8 \cdot 1 = x^8 \]
- For \(r = 1\):
\[ \binom{4}{1} (x^2)^{4-1} \left(\frac{2}{x}\right)^1 = 4 \cdot x^6 \cdot \frac{2}{x} = 8x^5 \]
- For \(r = 2\):
\[ \binom{4}{2} (x^2)^{4-2} \left(\frac{2}{x}\right)^2 = 6 \cdot x^4 \cdot \frac{4}{x^2} = 24x^2 \]
- For \(r = 3\):
\[ \binom{4}{3} (x^2)^{4-3} \left(\frac{2}{x}\right)^3 = 4 \cdot x^2 \cdot \frac{8}{x^3} = \frac{32}{x} \]
- For \(r = 4\):
\[ \binom{4}{4} (x^2)^{4-4} \left(\frac{2}{x}\right)^4 = 1 \cdot 1 \cdot \frac{16}{x^4} = \frac{16}{x^4} \]

4. **Combine all terms**:
\[ (x^2 + \frac{2}{x})^4 = x^8 + 8x^5 + 24x^2 + \frac{32}{x} + \frac{16}{x^4} \]

### Final Answers:
1. \((x + \frac{1}{x})^7 = x^7 + 7x^5 + 21x^3 + 35x + \frac{35}{x} + \frac{21}{x^3} + \frac{7}{x^5} + \frac{1}{x^7}\)
2. \((x^2 + \frac{2}{x})^4 = x^8 + 8x^5 + 24x^2 + \frac{32}{x} + \frac{16}{x^4}\)