(a) Use the binomial expansion formula: \((1 + ax)^n = 1 + nax + \frac{n(n-1)}{2!}a^2x^2 + \ldots\)

For \((1 + 2x)^5\):

First term: \(1\)

Second term: \(5 \times 2x = 10x\)

Third term: \(\frac{5 \times 4}{2} \times (2x)^2 = 40x^2\)

So, the first three terms are \(1 + 10x + 40x^2\).

(b) For \((1 - 3x)^4\):

First term: \(1\)

Second term: \(-4 \times 3x = -12x\)

Third term: \(\frac{4 \times 3}{2} \times (3x)^2 = 54x^2\)

So, the first three terms are \(1 - 12x + 54x^2\).

(c) To find the coefficient of \(x^2\) in \((1 + 2x)^5(1 - 3x)^4\), consider the products that result in \(x^2\):

1. \(1 \times 54x^2 = 54x^2\)

2. \(10x \times -12x = -120x^2\)

3. \(40x^2 \times 1 = 40x^2\)

Adding these gives: \(54 - 120 + 40 = -26\)

Thus, the coefficient of \(x^2\) is \(-26\).

(a) Use the binomial expansion for \(\left( 1 - \frac{1}{2x} \right)^2\):

\(= 1 - 2 \times \frac{1}{2x} + \left( \frac{1}{2x} \right)^2\)

\(= 1 - \frac{1}{x} + \frac{1}{4x^2}\).

(b) Use the binomial expansion for \((1 + 2x)^6\):

First four terms are:

\(1 + \binom{6}{1} (2x) + \binom{6}{2} (2x)^2 + \binom{6}{3} (2x)^3\)

\(= 1 + 12x + 60x^2 + 160x^3\).

(c) Multiply the expansions from (a) and (b) and find the coefficient of \(x\):

\((1 - \frac{1}{x} + \frac{1}{4x^2})(1 + 12x + 60x^2 + 160x^3)\)

Coefficient of \(x\) is:

\(1 \times 12 + (-1) \times 60 + \frac{1}{4} \times 160\)

\(= 12 - 60 + 40 = -8\).

(a) The binomial expansion of \((a-x)^6\) is given by:

\((a-x)^6 = a^6 - 6a^5x + 15a^4x^2 - 20a^3x^3 + \ldots\)

Thus, the first four terms are \(a^6 - 6a^5x + 15a^4x^2 - 20a^3x^3\).

(b) Consider the expansion of \(\left(1 + \frac{2}{ax}\right)(a-x)^6\).

The term in \(x^2\) arises from:

\(\left(1 + \frac{2}{ax}\right)(15a^4x^2 - 40a^3x^3 + \ldots)\)

The coefficient of \(x^2\) is \(15a^4 - 40a^2\).

Given that this coefficient is \(-20\), we have:

\(15a^4 - 40a^2 = -20\)

Rearranging gives:

\(15a^4 - 40a^2 + 20 = 0\)

Factorizing gives:

\((5a^2 - 10)(3a^2 - 2) = 0\)

Solving these equations gives:

\(a^2 = 2\) or \(a^2 = \frac{2}{3}\)

Thus, \(a = \pm \sqrt{2}\) or \(a = \pm \frac{\sqrt{2}}{3}\).

(a) To expand \((3 - 2x)^5\), use the binomial theorem:

\((3 - 2x)^5 = \sum_{k=0}^{5} \binom{5}{k} (3)^{5-k} (-2x)^k\).

The first three terms are:

\(\binom{5}{0} (3)^5 (-2x)^0 = 243\)

\(\binom{5}{1} (3)^4 (-2x)^1 = -810x\)

\(\binom{5}{2} (3)^3 (-2x)^2 = 1080x^2\)

Thus, the first three terms are \(243 - 810x + 1080x^2\).

(b) Expand \((4 + x)^2\):

\((4 + x)^2 = 16 + 8x + x^2\).

To find the coefficient of \(x^2\) in \((4 + x)^2(3 - 2x)^5\), consider:

\(16 \times 1080 + 8 \times (-810) + 243\).

Calculate:

\(16 \times 1080 = 17280\)

\(8 \times (-810) = -6480\)

\(17280 - 6480 + 243 = 11043\)

Thus, the coefficient of \(x^2\) is 11043.

(a) The binomial expansion of \((1 + x)^5\) is given by:

\(1 + 5x + \frac{5 \times 4}{2!}x^2 = 1 + 5x + 10x^2\).

(b) The binomial expansion of \((1 - 2x)^6\) is given by:

\(1 - 6(2x) + \frac{6 \times 5}{2!}(2x)^2 = 1 - 12x + 60x^2\).

(c) To find the coefficient of \(x^2\) in \((1 + x)^5 (1 - 2x)^6\), consider the products that result in \(x^2\):

- \(10x^2\) from \((1 + 5x + 10x^2) \times 1\)

- \(-60x^2\) from \(1 \times 60x^2\)

- \(60x^2\) from \(5x \times -12x\)

The coefficient of \(x^2\) is:

\(10 - 60 + 60 = 10\).