Expand using binomial theorem:
`(i) (1-2x)^(4) " " (ii) (x+2y)^(5)`
`(iii) (x-(1)/(x))^(6) " "(iv) ((2x)/(3)=(3)/(2x))^(5)`
`(v) (x^(2) +(2)/(x))^(6)" "(vi) (1+(1)/(x^(2)))^(4)`

Explanation

To expand the given expressions using the Binomial Theorem, we will follow the formula:

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

where \(\binom{n}{r}\) is the binomial coefficient, \(a\) and \(b\) are the terms being raised to the power \(n\), and \(r\) is the term index.

### (i) Expand \((1 - 2x)^4\)

1. Identify \(a = 1\), \(b = -2x\), and \(n = 4\).
2. Apply the Binomial Theorem:
\[ (1 - 2x)^4 = \sum_{r=0}^{4} \binom{4}{r} (1)^{4-r} (-2x)^r \]
3. Calculate each term:
- For \(r=0\): \(\binom{4}{0} (1)^4 (-2x)^0 = 1\)
- For \(r=1\): \(\binom{4}{1} (1)^3 (-2x)^1 = -8x\)
- For \(r=2\): \(\binom{4}{2} (1)^2 (-2x)^2 = 24x^2\)
- For \(r=3\): \(\binom{4}{3} (1)^1 (-2x)^3 = -32x^3\)
- For \(r=4\): \(\binom{4}{4} (1)^0 (-2x)^4 = 16x^4\)

4. Combine all terms:
\[ (1 - 2x)^4 = 1 - 8x + 24x^2 - 32x^3 + 16x^4 \]

### (ii) Expand \((x + 2y)^5\)

1. Identify \(a = x\), \(b = 2y\), and \(n = 5\).
2. Apply the Binomial Theorem:
\[ (x + 2y)^5 = \sum_{r=0}^{5} \binom{5}{r} (x)^{5-r} (2y)^r \]
3. Calculate each term:
- For \(r=0\): \(\binom{5}{0} x^5 (2y)^0 = x^5\)
- For \(r=1\): \(\binom{5}{1} x^4 (2y)^1 = 10x^4y\)
- For \(r=2\): \(\binom{5}{2} x^3 (2y)^2 = 40x^3y^2\)
- For \(r=3\): \(\binom{5}{3} x^2 (2y)^3 = 80x^2y^3\)
- For \(r=4\): \(\binom{5}{4} x^1 (2y)^4 = 80xy^4\)
- For \(r=5\): \(\binom{5}{5} x^0 (2y)^5 = 32y^5\)

4. Combine all terms:
\[ (x + 2y)^5 = x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5 \]

### (iii) Expand \((x - \frac{1}{x})^6\)

1. Identify \(a = x\), \(b = -\frac{1}{x}\), and \(n = 6\).
2. Apply the Binomial Theorem:
\[ (x - \frac{1}{x})^6 = \sum_{r=0}^{6} \binom{6}{r} (x)^{6-r} \left(-\frac{1}{x}\right)^r \]
3. Calculate each term:
- For \(r=0\): \(\binom{6}{0} x^6 = x^6\)
- For \(r=1\): \(\binom{6}{1} x^5 \left(-\frac{1}{x}\right) = -6x^4\)
- For \(r=2\): \(\binom{6}{2} x^4 \left(-\frac{1}{x}\right)^2 = 15x^2\)
- For \(r=3\): \(\binom{6}{3} x^3 \left(-\frac{1}{x}\right)^3 = -20\)
- For \(r=4\): \(\binom{6}{4} x^2 \left(-\frac{1}{x}\right)^4 = 15\frac{1}{x^2}\)
- For \(r=5\): \(\binom{6}{5} x^1 \left(-\frac{1}{x}\right)^5 = -6\frac{1}{x^4}\)
- For \(r=6\): \(\binom{6}{6} \left(-\frac{1}{x}\right)^6 = \frac{1}{x^6}\)

4. Combine all terms:
\[ (x - \frac{1}{x})^6 = x^6 - 6x^4 + 15x^2 - 20 + 15\frac{1}{x^2} - 6\frac{1}{x^4} + \frac{1}{x^6} \]

### (iv) Expand \(\left(\frac{2x}{3} + \frac{3}{2x}\right)^5\)

1. Identify \(a = \frac{2x}{3}\), \(b = \frac{3}{2x}\), and \(n = 5\).
2. Apply the Binomial Theorem:
\[ \left(\frac{2x}{3} + \frac{3}{2x}\right)^5 = \sum_{r=0}^{5} \binom{5}{r} \left(\frac{2x}{3}\right)^{5-r} \left(\frac{3}{2x}\right)^r \]
3. Calculate each term:
- For \(r=0\): \(\binom{5}{0} \left(\frac{2x}{3}\right)^5 = \frac{32x^5}{243}\)
- For \(r=1\): \(\binom{5}{1} \left(\frac{2x}{3}\right)^4 \left(\frac{3}{2x}\right) = \frac{5 \cdot 16x^4 \cdot 3}{3 \cdot 16} = \frac{80x^3}{81}\)
- For \(r=2\): \(\binom{5}{2} \left(\frac{2x}{3}\right)^3 \left(\frac{3}{2x}\right)^2 = \frac{5 \cdot 8x^3 \cdot 9}{9 \cdot 4} = \frac{90x^2}{27} = \frac{10x^2}{3}\)
- For \(r=3\): \(\binom{5}{3} \left(\frac{2x}{3}\right)^2 \left(\frac{3}{2x}\right)^3 = \frac{10 \cdot 4x^2 \cdot 27}{27 \cdot 8} = \frac{10}{6} = \frac{5}{3}\)
- For \(r=4\): \(\binom{5}{4} \left(\frac{2x}{3}\right)^1 \left(\frac{3}{2x}\right)^4 = \frac{5 \cdot 2x \cdot 81}{3 \cdot 16} = \frac{10}{16} = \frac{5}{8}\)
- For \(r=5\): \(\binom{5}{5} \left(\frac{3}{2x}\right)^5 = \frac{243}{32x^5}\)

4. Combine all terms:
\[ \left(\frac{2x}{3} + \frac{3}{2x}\right)^5 = \frac{32x^5}{243} + \frac{80x^3}{81} + \frac{10x^2}{3} + \frac{5}{3} + \frac{5}{8} + \frac{243}{32x^5} \]

### (v) Expand \((x^2 + \frac{2}{x})^6\)

1. Identify \(a = x^2\), \(b = \frac{2}{x}\), and \(n = 6\).
2. Apply the Binomial Theorem:
\[ (x^2 + \frac{2}{x})^6 = \sum_{r=0}^{6} \binom{6}{r} (x^2)^{6-r} \left(\frac{2}{x}\right)^r \]
3. Calculate each term:
- For \(r=0\): \(\binom{6}{0} (x^2)^6 = x^{12}\)
- For \(r=1\): \(\binom{6}{1} (x^2)^5 \left(\frac{2}{x}\right) = 12x^{10}\)
- For \(r=2\): \(\binom{6}{2} (x^2)^4 \left(\frac{2}{x}\right)^2 = 30x^8 \cdot 4 = 120x^8\)
- For \(r=3\): \(\binom{6}{3} (x^2)^3 \left(\frac{2}{x}\right)^3 = 20x^6 \cdot 8 = 160x^6\)
- For \(r=4\): \(\binom{6}{4} (x^2)^2 \left(\frac{2}{x}\right)^4 = 15x^4 \cdot 16 = 240x^4\)
- For \(r=5\): \(\binom{6}{5} (x^2)^1 \left(\frac{2}{x}\right)^5 = 6x^2 \cdot 32 = 192x^2\)
- For \(r=6\): \(\binom{6}{6} \left(\frac{2}{x}\right)^6 = \frac{64}{x^6}\)

4. Combine all terms:
\[ (x^2 + \frac{2}{x})^6 = x^{12} + 12x^{10} + 120x^8 + 160x^6 + 240x^4 + 192x^2 + \frac{64}{x^6} \]

### (vi) Expand \((1 + \frac{1}{x^2})^4\)

1. Identify \(a = 1\), \(b = \frac{1}{x^2}\), and \(n = 4\).
2. Apply the Binomial Theorem:
\[ (1 + \frac{1}{x^2})^4 = \sum_{r=0}^{4} \binom{4}{r} (1)^{4-r} \left(\frac{1}{x^2}\right)^r \]
3. Calculate each term:
- For \(r=0\): \(\binom{4}{0} (1)^4 = 1\)
- For \(r=1\): \(\binom{4}{1} (1)^3 \left(\frac{1}{x^2}\right) = \frac{4}{x^2}\)
- For \(r=2\): \(\binom{4}{2} (1)^2 \left(\frac{1}{x^2}\right)^2 = 6\frac{1}{x^4}\)
- For \(r=3\): \(\binom{4}{3} (1)^1 \left(\frac{1}{x^2}\right)^3 = \frac{4}{x^6}\)
- For \(r=4\): \(\binom{4}{4} \left(\frac{1}{x^2}\right)^4 = \frac{1}{x^8}\)

4. Combine all terms:
\[ (1 + \frac{1}{x^2})^4 = 1 + \frac{4}{x^2} + \frac{6}{x^4} + \frac{4}{x^6} + \frac{1}{x^8} \]