(a)(i) Expand \((1 + 2x)^5\) using the binomial theorem:

\((1 + 2x)^5 = 1 + 5(2x) + \frac{5 \times 4}{2}(2x)^2 = 1 + 10x + 40x^2\).

(a)(ii) Expand \((1 - ax)^6\) using the binomial theorem:

\((1 - ax)^6 = 1 - 6(ax) + \frac{6 \times 5}{2}(ax)^2 = 1 - 6ax + 15a^2x^2\).

(b) To find the coefficient of \(x^2\) in \((1 + 2x)^5(1 - ax)^6\), consider the terms that contribute to \(x^2\):

1. \(x^2\) term from \((1 + 2x)^5\) and constant term from \((1 - ax)^6\): \(40x^2 \times 1 = 40x^2\).

2. \(x\) term from \((1 + 2x)^5\) and \(x\) term from \((1 - ax)^6\): \(10x \times (-6ax) = -60ax^2\).

3. Constant term from \((1 + 2x)^5\) and \(x^2\) term from \((1 - ax)^6\): \(1 \times 15a^2x^2 = 15a^2x^2\).

Combine these to find the total coefficient of \(x^2\):

\(40 - 60a + 15a^2 = -5\).

Simplify and solve the quadratic equation:

\(15a^2 - 60a + 40 = -5\).

\(15a^2 - 60a + 45 = 0\).

Divide by 15:

\(a^2 - 4a + 3 = 0\).

Factorize:

\((a - 1)(a - 3) = 0\).

Thus, \(a = 1\) and \(a = 3\).

(a) To find the first three terms in the expansion of \((1 + ax)^6\), we use the binomial theorem:

\((1 + ax)^6 = 1 + \binom{6}{1}(ax) + \binom{6}{2}(ax)^2 + \ldots\)

Calculating the first three terms:

\(1 + 6ax + 15a^2x^2\)

(b) To find the coefficient of \(x^2\) in the expansion of \((1 - 3x)(1 + ax)^6\), we first find the relevant terms from each factor:

The expansion of \((1 + ax)^6\) gives the terms \(1, 6ax, 15a^2x^2\).

We need the coefficient of \(x^2\) from the product:

\((1)(15a^2x^2) + (-3x)(6ax) = 15a^2x^2 - 18ax^2\)

Setting the coefficient equal to \(-3\):

\(15a^2 - 18a = -3\)

Rearranging gives:

\(15a^2 - 18a + 3 = 0\)

Factoring the quadratic:

\(3(5a^2 - 6a + 1) = 0\)

\((3)(a-1)(5a-1) = 0\)

Thus, \(a = 1\) or \(a = \frac{1}{5}\).

(a) To find the coefficient of \(x^2\) in the expansion of \((4 + 2x)(2 - ax)^5\), consider the terms that contribute to \(x^2\):

- From \((4)(2 - ax)^5\), the term contributing to \(x^2\) is \(-80ax\).

- From \((2x)(2 - ax)^5\), the term contributing to \(x^2\) is \(80a^2x^2\).

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

\(320a^2 - 160a = -15\)

Rearranging gives:

\(320a^2 - 160a + 15 = 0\)

Solving this quadratic equation using the quadratic formula:

\(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 320\), \(b = -160\), \(c = 15\).

\(a = \frac{160 \pm \sqrt{(-160)^2 - 4 \times 320 \times 15}}{2 \times 320}\)

\(a = \frac{160 \pm \sqrt{25600 - 19200}}{640}\)

\(a = \frac{160 \pm \sqrt{6400}}{640}\)

\(a = \frac{160 \pm 80}{640}\)

\(a = \frac{240}{640} \text{ or } a = \frac{80}{640}\)

\(a = \frac{3}{8} \text{ or } a = \frac{1}{8}\)

(b) Given that there is only one value of \(a\) for which the coefficient of \(x^2\) is \(k\), we set the discriminant of the quadratic equation to zero:

\(160^2 - 4 \times 320 \times (15 - k) = 0\)

Solving for \(k\):

\(25600 - 1280(15 - k) = 0\)

\(25600 - 19200 + 1280k = 0\)

\(6400 + 1280k = 0\)

\(k = -\frac{6400}{1280}\)

\(k = -20\)

Substituting \(k = -20\) back into the quadratic equation gives:

\(320a^2 - 160a + 20 = 0\)

Solving this gives \(a = \frac{1}{4}\).

First, find the coefficient of \(x\) in the expansion of \(\left(4x + \frac{10}{x}\right)^3\).

The term that gives \(x\) is \(\binom{3}{2} (4x)^2 \left(\frac{10}{x}\right)^1 = 3 \times 16x^2 \times \frac{10}{x} = 480x\).

Thus, \(p = 480\).

Next, find the coefficient of \(\frac{1}{x}\) in the expansion of \(\left(2x + \frac{k}{x^2}\right)^5\).

The term that gives \(\frac{1}{x}\) is \(\binom{5}{3} (2x)^3 \left(\frac{k}{x^2}\right)^2 = 10 \times 8x^3 \times \frac{k^2}{x^4} = 80k^2 \times \frac{1}{x}\).

Thus, \(q = 80k^2\).

Given \(p = 6q\), we have:

\(480 = 6 \times 80k^2\)

\(480 = 480k^2\)

\(k^2 = 1\)

Therefore, \(k = \pm 1\).

First, find the coefficient of \(x^3\) in \((1 - 2x)^5\). The general term in the binomial expansion of \((1 - 2x)^5\) is given by:

\(\binom{5}{r} (1)^{5-r} (-2x)^r = \binom{5}{r} (-2)^r x^r\).

For \(x^3\), set \(r = 3\):

\(\binom{5}{3} (-2)^3 x^3 = 10(-8)x^3 = -80x^3\).

Thus, the coefficient of \(x^3\) in \((1 - 2x)^5\) is \(-80\).

Next, find the coefficient of \(x^2\) in \((1 - 2x)^5\) to combine with \(kx\) from \((1 + kx)\):

For \(x^2\), set \(r = 2\):

\(\binom{5}{2} (-2)^2 x^2 = 10(4)x^2 = 40x^2\).

Thus, the coefficient of \(x^2\) in \((1 - 2x)^5\) is \(40\).

The coefficient of \(x^3\) in \((1 + kx)(1 - 2x)^5\) is given by:

\(40k - 80 = 20\).

Solving for \(k\):

\(40k - 80 = 20\)

\(40k = 100\)

\(k = \frac{100}{40} = \frac{5}{2}\).

(a) Consider the expansion of \((2x^2 + \frac{a}{x})^6\). The general term is given by:

\(\binom{6}{r} (2x^2)^{6-r} \left(\frac{a}{x}\right)^r\)

For \(x^6\), we have:

\(\binom{6}{2} (2x^2)^4 \left(\frac{a}{x}\right)^2 = 15 \times 16x^8 \times \frac{a^2}{x^2} = 15 \times 16a^2 x^6\)

For \(x^3\), we have:

\(\binom{6}{3} (2x^2)^3 \left(\frac{a}{x}\right)^3 = 20 \times 8x^6 \times \frac{a^3}{x^3} = 20 \times 8a^3 x^3\)

Equating the coefficients:

\(15 \times 16a^2 = 20 \times 8a^3\)

\(240a^2 = 160a^3\)

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

(b) In the expansion of \((1-x^3)(2x^2 + \frac{a}{x})^6\), the term \(x^6\) comes from the product of \(-x^3\) and the constant term in \((2x^2 + \frac{a}{x})^6\), which is zero. Therefore, the coefficient of \(x^6\) is 0.

To find the coefficient of \(\frac{1}{x}\) in the expansion, we consider each part separately:

1. For \(\left( kx + \frac{1}{x} \right)^5\), the term that contributes to \(\frac{1}{x}\) is obtained by choosing 4 factors of \(kx\) and 1 factor of \(\frac{1}{x}\). The coefficient is:

\(\binom{5}{1} (kx)^4 \left( \frac{1}{x} \right)^1 = 5 \cdot k^4 \cdot x^3 \cdot \frac{1}{x} = 5k^4 x^2\)

2. For \(\left( 1 - \frac{2}{x} \right)^8\), the term that contributes to \(\frac{1}{x}\) is obtained by choosing 7 factors of 1 and 1 factor of \(-\frac{2}{x}\). The coefficient is:

\(\binom{8}{1} 1^7 \left( -\frac{2}{x} \right)^1 = 8 \cdot (-2) = -16\)

Combining these, the total coefficient of \(\frac{1}{x}\) is:

\(10k^2 - 16\)

We are given that this coefficient is 74:

\(10k^2 - 16 = 74\)

Solving for \(k\):

\(10k^2 = 90\)

\(k^2 = 9\)

\(k = 3\)

(a) To find the coefficient of \(\frac{1}{x}\), consider the term \(\binom{5}{2} (2x)^3 \left( \frac{a}{x^2} \right)^2\).

This simplifies to \(10 \times 8x^3 \times \frac{a^2}{x^4} = 720 \times \frac{1}{x}\).

Thus, \(80a^2 = 720\), giving \(a^2 = 9\).

Therefore, \(a = \pm 3\).

(b) To find the coefficient of \(\frac{1}{x^7}\), consider the term \(\binom{5}{4} (2x)^1 \left( \frac{a}{x^2} \right)^4\).

This simplifies to \(5 \times 2x \times \frac{a^4}{x^8}\).

Using \(a = 3\), the coefficient is \(5 \times 2 \times 81 = 810\).

First, expand \(\left(1 + \frac{x}{2}\right)^6\) using the binomial theorem:

\(\left(1 + \frac{x}{2}\right)^6 = 1 + \binom{6}{1}\frac{x}{2} + \binom{6}{2}\left(\frac{x}{2}\right)^2 + \cdots\)

The terms involving \(x\) and \(x^2\) are:

\(\binom{6}{1}\frac{x}{2} = 6\frac{x}{2} = 3x\)

\(\binom{6}{2}\left(\frac{x}{2}\right)^2 = 15\frac{x^2}{4}\)

Now, consider the expansion of \((4 + ax)(1 + 3x + 15\frac{x^2}{4})\).

The coefficient of \(x^2\) comes from:

\(4 \times 15\frac{x^2}{4} = 15x^2\)

\(ax \times 3x = 3ax^2\)

Equating the coefficient of \(x^2\) to 3:

\(3a + 15 = 3\)

Solving for \(a\):

\(3a = 3 - 15\)

\(3a = -12\)

\(a = -4\)

(i) To find the term independent of x, consider the general term in the expansion:

\(T_r = \binom{6}{r} (2x)^{6-r} \left( \frac{k}{x} \right)^r\)

The term independent of x occurs when the powers of x cancel out, i.e.,

\((6-r) - r = 0\)

\(6 - 2r = 0\)

\(r = 3\)

Substitute \(r = 3\) into the term:

\(T_3 = \binom{6}{3} (2x)^3 \left( \frac{k}{x} \right)^3\)

\(= 20 \cdot 8x^3 \cdot \frac{k^3}{x^3}\)

\(= 20 \cdot 8 \cdot k^3\)

\(= 160k^3\)

Given that this term is 540, we have:

\(160k^3 = 540\)

\(k^3 = \frac{540}{160}\)

\(k^3 = \frac{27}{8}\)

\(k = \frac{1}{2}\)

(ii) To find the coefficient of x², consider the term:

\(T_r = \binom{6}{r} (2x)^{6-r} \left( \frac{k}{x} \right)^r\)

We need \((6-r) - r = 2\), so:

\(6 - 2r = 2\)

\(4 = 2r\)

\(r = 2\)

Substitute \(r = 2\) into the term:

\(T_2 = \binom{6}{2} (2x)^4 \left( \frac{k}{x} \right)^2\)

\(= 15 \cdot 16x^4 \cdot \frac{k^2}{x^2}\)

\(= 15 \cdot 16 \cdot k^2 \cdot x^2\)

Substitute \(k = \frac{1}{2}\):

\(= 15 \cdot 16 \cdot \left( \frac{1}{2} \right)^2\)

\(= 15 \cdot 16 \cdot \frac{1}{4}\)

\(= 240\)

To find the coefficient of \(x^3\) in the expansion of \((1 - px)^5\), we use the binomial theorem:

\((1 - px)^5 = \sum_{k=0}^{5} \binom{5}{k} (1)^{5-k} (-px)^k\).

The term containing \(x^3\) is when \(k = 3\):

\(\binom{5}{3} (1)^{5-3} (-px)^3 = \binom{5}{3} (-1)^3 p^3 x^3\).

\(\binom{5}{3} = 10\), so the coefficient of \(x^3\) is:

\((-1) \cdot 10 \cdot p^3 = -10p^3\).

We are given that this coefficient is \(-2160\):

\(-10p^3 = -2160\).

Dividing both sides by \(-10\):

\(p^3 = 216\).

Taking the cube root of both sides:

\(p = 6\).

The coefficient of \(x^3\) in the expansion of \((3 + 2ax)^5\) is given by:

\(\binom{5}{3} (3)^{2} (2ax)^{3} = 10 \times 9 \times 8a^3 = 720a^3\)

The coefficient of \(x^2\) in the expansion of \((2 + ax)^6\) is given by:

\(\binom{6}{2} (2)^{4} (ax)^{2} = 15 \times 16 \times a^2 = 240a^2\)

According to the problem, the coefficient of \(x^3\) is six times the coefficient of \(x^2\):

\(720a^3 = 6 \times 240a^2\)

Simplifying gives:

\(720a^3 = 1440a^2\)

Dividing both sides by \(240a^2\) gives:

\(3a = 6\)

Thus, \(a = 2\).

First, find the coefficient of \(x^2\) in \(\left( 2 + \frac{x}{2} \right)^6\).

Using the binomial theorem, the term containing \(x^2\) is given by:

\(\binom{6}{2} \cdot 2^4 \cdot \left( \frac{x}{2} \right)^2 = 15 \cdot 16 \cdot \frac{x^2}{4} = 60x^2\)

So, the coefficient of \(x^2\) in \(\left( 2 + \frac{x}{2} \right)^6\) is 60.

Next, find the coefficient of \(x^2\) in \((a + x)^5\).

Using the binomial theorem, the term containing \(x^2\) is given by:

\(\binom{5}{2} \cdot a^3 \cdot x^2 = 10a^3x^2\)

So, the coefficient of \(x^2\) in \((a + x)^5\) is \(10a^3\).

According to the problem, the sum of these coefficients is 330:

\(60 + 10a^3 = 330\)

Solve for \(a^3\):

\(10a^3 = 270\)

\(a^3 = 27\)

\(a = 3\)

To find the coefficients of x and x² in the expansion of (2 + ax)⁷, we use the binomial theorem.

The general term in the expansion is given by:

\(\binom{7}{r} (2)^{7-r} (ax)^r\)

\(For the coefficient of x, set r = 1:\)

\(\binom{7}{1} (2)^6 (ax) = 7 \times 2^6 \times a \times x\)

The coefficient is \(7 \times 2^6 \times a\).

\(For the coefficient of x², set r = 2:\)

\(\binom{7}{2} (2)^5 (ax)^2 = 21 \times 2^5 \times a^2 \times x^2\)

The coefficient is \(21 \times 2^5 \times a^2\).

Since the coefficients are equal, we have:

\(7 \times 2^6 \times a = 21 \times 2^5 \times a^2\)

Simplifying gives:

\(7 \times 2 = 21 \times a\)

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

To find the coefficients of x² and x³ in the expansion of (3 - 2x)⁶, we use the binomial theorem:

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

For the term involving x², we have:

\(a = \binom{6}{2} \cdot 3^{4} \cdot (-2x)^{2}\)

\(a = 15 \cdot 81 \cdot 4\)

\(a = 4860\)

For the term involving x³, we have:

\(b = \binom{6}{3} \cdot 3^{3} \cdot (-2x)^{3}\)

\(b = 20 \cdot 27 \cdot (-8)\)

\(b = -4320\)

Thus, \(\frac{a}{b} = \frac{4860}{-4320} = -\frac{9}{8}\).

Therefore, the value of \(\frac{a}{b}\) is \(\frac{9}{8}\).

To find the coefficient of \(x\) in the expansion of \(\left( \frac{1}{ax} + 2ax^2 \right)^5\), we use the binomial theorem. The general term in the expansion is given by:

\(T_k = \binom{5}{k} \left( \frac{1}{ax} \right)^{5-k} (2ax^2)^k\)

We need the term where the power of \(x\) is 1. The power of \(x\) in the general term is:

\(-5 + k + 2k = -5 + 3k\)

Setting \(-5 + 3k = 1\), we solve for \(k\):

\(3k = 6\)

\(k = 2\)

Substitute \(k = 2\) into the general term:

\(T_2 = \binom{5}{2} \left( \frac{1}{ax} \right)^{3} (2ax^2)^2\)

\(= 10 \times \frac{1}{a^3x^3} \times 4a^2x^4\)

\(= 10 \times \frac{4a^2x^4}{a^3x^3}\)

\(= 10 \times \frac{4a^2x}{a^3}\)

\(= \frac{40a^2x}{a^3}\)

\(= \frac{40x}{a}\)

The coefficient of \(x\) is \(\frac{40}{a}\). We are given that this coefficient is 5:

\(\frac{40}{a} = 5\)

Solving for \(a\):

\(40 = 5a\)

\(a = 8\)

To find the coefficient of \(x^3\) in the expansion of \((1 - 3x)^6 + (1 + ax)^5\), we consider the contributions from each binomial expansion separately.

For \((1 - 3x)^6\), the term containing \(x^3\) is given by the binomial coefficient:

\(\binom{6}{3} (1)^{6-3} (-3x)^3 = 20(-27x^3) = -540x^3.\)

For \((1 + ax)^5\), the term containing \(x^3\) is:

\(\binom{5}{3} (1)^{5-3} (ax)^3 = 10a^3x^3.\)

The total coefficient of \(x^3\) is:

\(-540 + 10a^3 = 100.\)

Solving for \(a\), we have:

\(10a^3 = 640\)

\(a^3 = 64\)

\(a = 4.\)

The term in \(x\) from \(\left(1 + \frac{x}{2}\right)^n\) is \(\frac{nx}{2}\).

In the expansion of \((3 - 2x)(1 + \frac{nx}{2} + \ldots)\), the coefficient of \(x\) is given by:

\(3 \cdot \frac{nx}{2} - 2 = 7\)

\(\frac{3n}{2} - 2 = 7\)

\(\frac{3n}{2} = 9\)

\(n = 6\)

Now, find the coefficient of \(x^2\):

The term in \(x^2\) from \(\left(1 + \frac{x}{2}\right)^n\) is \(\frac{n(n-1)}{2} \left(\frac{x}{2}\right)^2\).

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

\(\frac{3n(n-1)}{8} - \frac{2n}{2}\)

Substitute \(n = 6\):

\(\frac{3 \times 6 \times 5}{8} - 6 = \frac{90}{8} - 6 = \frac{45}{4} - 6 = \frac{21}{4}\)

The general term in the expansion of \((x + 2k)^7\) is given by:

\(T_r = \binom{7}{r} (x)^{7-r} (2k)^r\).

For the term in \(x^5\), set \(7-r = 5\), so \(r = 2\):

\(T_5 = \binom{7}{2} x^5 (2k)^2 = 21 \cdot 4k^2 x^5 = 84k^2 x^5\).

For the term in \(x^4\), set \(7-r = 4\), so \(r = 3\):

\(T_4 = \binom{7}{3} x^4 (2k)^3 = 35 \cdot 8k^3 x^4 = 280k^3 x^4\).

Since the coefficients of \(x^4\) and \(x^5\) are equal, we equate them:

\(84k^2 = 280k^3\).

Divide both sides by \(k^2\) (since \(k\) is non-zero):

\(84 = 280k\).

Solve for \(k\):

\(k = \frac{84}{280} = \frac{3}{10} = 0.3\).

To find the coefficient of \(x^2\) in the expansion of \((2 + ax)^6\), we use the binomial theorem:

\(\binom{6}{2} \cdot 2^4 \cdot (ax)^2 = 15 \cdot 16 \cdot a^2 x^2 = 240a^2 x^2\)

To find the coefficient of \(x^3\), we have:

\(\binom{6}{3} \cdot 2^3 \cdot (ax)^3 = 20 \cdot 8 \cdot a^3 x^3 = 160a^3 x^3\)

Setting the coefficients equal gives:

\(240a^2 = 160a^3\)

Solving for \(a\), we divide both sides by \(a^2\) (assuming \(a \neq 0\)):

\(240 = 160a\)

\(a = \frac{240}{160} = \frac{3}{2}\)

The general term in the expansion of \((2 + ax)^7\) is given by:

\(T_k = \binom{7}{k} (2)^{7-k} (ax)^k\).

For the coefficient of \(x\), set \(k = 1\):

\(T_1 = \binom{7}{1} (2)^6 (ax)^1 = 7 \times 2^6 \times a \times x\).

The coefficient of \(x\) is \(7 \times 2^6 \times a\).

For the coefficient of \(x^2\), set \(k = 2\):

\(T_2 = \binom{7}{2} (2)^5 (ax)^2 = 21 \times 2^5 \times a^2\).

The coefficient of \(x^2\) is \(21 \times 2^5 \times a^2\).

Equating the coefficients:

\(7 \times 2^6 \times a = 21 \times 2^5 \times a^2\).

Divide both sides by \(2^5\):

\(7 \times 2 \times a = 21 \times a^2\).

Divide both sides by \(a\) (since \(a \neq 0\)):

\(14 = 21a\).

Solve for \(a\):

\(a = \frac{14}{21} = \frac{2}{3}\).

To find the coefficient of \(x^5\) in the expansion of \(\left(x^2 - \frac{a}{x}\right)^7\), we use the binomial theorem.

The general term in the expansion is given by:

\(T_k = \binom{7}{k} (x^2)^{7-k} \left(-\frac{a}{x}\right)^k\)

We need the term where the power of \(x\) is 5:

\((x^2)^{7-k} \left(-\frac{a}{x}\right)^k = x^{2(7-k)} \cdot (-a)^k \cdot x^{-k} = x^{14-2k-k} = x^{14-3k}\)

Set the exponent equal to 5:

\(14 - 3k = 5\)

Solving for \(k\):

\(14 - 5 = 3k\)

\(9 = 3k\)

\(k = 3\)

Substitute \(k = 3\) into the general term:

\(T_3 = \binom{7}{3} (x^2)^4 \left(-\frac{a}{x}\right)^3\)

\(= \binom{7}{3} x^8 \cdot (-a)^3 \cdot x^{-3}\)

\(= \binom{7}{3} (-a)^3 x^5\)

The coefficient of \(x^5\) is:

\(\binom{7}{3} (-a)^3 = -280\)

Calculate \(\binom{7}{3}\):

\(\binom{7}{3} = 35\)

Thus:

\(35(-a)^3 = -280\)

\((-a)^3 = -8\)

\(a^3 = 8\)

\(a = 2\)

(a) To expand \(\left( x + \frac{2}{x} \right)^5\), use the binomial theorem:

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

Calculate each term:

\(\binom{5}{0} x^5 = x^5\)

\(\binom{5}{1} x^4 \cdot \frac{2}{x} = 10x^3\)

\(\binom{5}{2} x^3 \cdot \left( \frac{2}{x} \right)^2 = 40x\)

\(\binom{5}{3} x^2 \cdot \left( \frac{2}{x} \right)^3 = \frac{80}{x}\)

\(\binom{5}{4} x \cdot \left( \frac{2}{x} \right)^4 = \frac{80}{x^3}\)

\(\binom{5}{5} \left( \frac{2}{x} \right)^5 = \frac{32}{x^5}\)

Thus, the expansion is \(x^5 + 10x^3 + 40x + \frac{80}{x} + \frac{80}{x^3} + \frac{32}{x^5}\).

(b) Consider the expansion \((a + bx^2) \left( x + \frac{2}{x} \right)^5\).

The coefficient of \(x\) is zero:

\(40a + (\text{coefficient of } x^{-1})b = 0\)

The coefficient of \(\frac{1}{x}\) is 80:

\((\text{coefficient of } x^{-1})a + (\text{coefficient of } x^{-3})b = 80\)

From part (a), the coefficient of \(x^{-1}\) is 80 and \(x^{-3}\) is 80.

Thus, the equations are:

\(40a + 80b = 0\)

\(80a + 80b = 80\)

Solving these equations:

From \(40a + 80b = 0\), we get \(a = -2b\).

Substitute into \(80a + 80b = 80\):

\(80(-2b) + 80b = 80\)

\(-160b + 80b = 80\)

\(-80b = 80\)

\(b = -1\)

Substitute \(b = -1\) into \(a = -2b\):

\(a = 2\)

Thus, \(a = 2\) and \(b = -1\).

To find the coefficient of \(x^3\) in the expansion of \((a + x)^5\), we use the binomial theorem:

\(\text{Coeff of } x^3 \text{ in } (a + x)^5 = \binom{5}{3} a^{5-3} x^3 = 10a^2\)

For \((2 - x)^6\), the coefficient of \(x^3\) is:

\(\text{Coeff of } x^3 \text{ in } (2 - x)^6 = \binom{6}{3} (2)^{6-3} (-1)^3 x^3 = -160\)

Adding these coefficients gives the total coefficient of \(x^3\):

\(10a^2 - 160 = 90\)

Solving for \(a\):

\(10a^2 = 250\)

\(a^2 = 25\)

\(a = 5\)

To find the coefficient of \(x^2\) in the expansion of \(\left( k + \frac{1}{3}x \right)^5\), we use the binomial theorem:

\(\binom{5}{2} k^{5-2} \left( \frac{1}{3}x \right)^2\)

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

\(\binom{5}{2} k^3 \left( \frac{1}{3} \right)^2\)

\(= 10 \times k^3 \times \frac{1}{9}\)

We know this coefficient is 30, so:

\(10 \times k^3 \times \frac{1}{9} = 30\)

\(\frac{10}{9} k^3 = 30\)

\(k^3 = 30 \times \frac{9}{10}\)

\(k^3 = 27\)

\(k = 3\)

To find the coefficient of \(x^3\) in the expansion of \((a+x)^5 + (1-2x)^6\), we need to consider the individual expansions.

For \((a+x)^5\), the term containing \(x^3\) is given by the binomial coefficient:

\(\binom{5}{3} a^2 x^3 = 10a^2 x^3\)

For \((1-2x)^6\), the term containing \(x^3\) is:

\(\binom{6}{3} (1)^3 (-2x)^3 = 20(-8)x^3 = -160x^3\)

The total coefficient of \(x^3\) in the expansion is:

\(10a^2 - 160 = 90\)

Solving for \(a\):

\(10a^2 = 250\)

\(a^2 = 25\)

\(a = 5\)

The coefficient of \(x^4\) in the expansion of \((x + a)^6\) is given by \(\binom{6}{4} a^2 = 15a^2\). Thus, \(p = 15a^2\).

The coefficient of \(x^2\) in the expansion of \((ax + 3)^4\) is given by \(\binom{4}{2} (ax)^2 3^2 = 54a^2\). Thus, \(q = 54a^2\).

We know that \(p + q = 276\), so:

\(15a^2 + 54a^2 = 276\)

\(69a^2 = 276\)

\(a^2 = \frac{276}{69} = 4\)

\(a = \pm 2\)

Consider the binomial expansion of \(\left( \frac{x}{a} + \frac{a}{x^2} \right)^7\).

The general term is given by:

\(T_k = \binom{7}{k} \left( \frac{x}{a} \right)^{7-k} \left( \frac{a}{x^2} \right)^k\)

Simplifying, we have:

\(T_k = \binom{7}{k} \frac{x^{7-k}}{a^{7-k}} \frac{a^k}{x^{2k}} = \binom{7}{k} \frac{a^{k-(7-k)}}{x^{2k-(7-k)}} = \binom{7}{k} \frac{a^{2k-7}}{x^{k+7-2k}}\)

For the coefficient of \(x^4\), set \(k+7-2k = 4\), giving \(k = 3\).

The coefficient of \(x^4\) is:

\(\binom{7}{3} \frac{a^{6-7}}{x^4} = \frac{35}{a}\)

For the coefficient of \(x\), set \(k+7-2k = 1\), giving \(k = 6\).

The coefficient of \(x\) is:

\(\binom{7}{6} \frac{a^{12-7}}{x} = 7a^5\)

Given:

\(\frac{\frac{35}{a}}{7a^5} = 3\)

Simplifying:

\(\frac{5}{a^6} = 3\)

\(a^6 = \frac{5}{3}\)

\(a^2 = \frac{1}{9}\)

\(a = \pm \frac{1}{3}\)

The coefficient of \(x^2\) in \(\left( 1 + \frac{2}{p} x \right)^5\) is given by:

\(\binom{5}{2} \left( \frac{2}{p} \right)^2 = 10 \times \frac{4}{p^2} = \frac{40}{p^2}\)

The coefficient of \(x^2\) in \((1 + px)^6\) is given by:

\(\binom{6}{2} (p)^2 = 15p^2\)

Setting the sum of these coefficients equal to 70:

\(\frac{40}{p^2} + 15p^2 = 70\)

Multiply through by \(p^2\) to clear the fraction:

\(40 + 15p^4 = 70p^2\)

Rearrange to form a quadratic in \(p^2\):

\(15p^4 - 70p^2 + 40 = 0\)

Let \(q = p^2\), then:

\(15q^2 - 70q + 40 = 0\)

Using the quadratic formula \(q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\(q = \frac{70 \pm \sqrt{70^2 - 4 \times 15 \times 40}}{30}\)

\(q = \frac{70 \pm \sqrt{4900 - 2400}}{30}\)

\(q = \frac{70 \pm \sqrt{2500}}{30}\)

\(q = \frac{70 \pm 50}{30}\)

This gives \(q = 4\) or \(q = \frac{2}{3}\).

Thus, \(p^2 = 4\) or \(p^2 = \frac{2}{3}\).

Therefore, \(p = \pm 2\) or \(p = \pm \frac{\sqrt{2}}{3}\).

To find the coefficient of \(x^3\) in the expansion of \(\left(p + \frac{1}{p}x\right)^4\), we use the binomial theorem:

\(\binom{4}{3} \cdot p^{4-3} \cdot \left(\frac{1}{p}\right)^3 \cdot x^3 = 4 \cdot p \cdot \frac{1}{p^3} \cdot x^3 = \frac{4}{p^2}x^3\).

The coefficient of \(x^3\) is \(\frac{4}{p^2}\).

We are given that this coefficient is 144:

\(\frac{4}{p^2} = 144\).

Solving for \(p^2\), we get:

\(p^2 = \frac{4}{144} = \frac{1}{36}\).

Taking the square root of both sides, we find:

\(p = \pm \frac{1}{6}\).

First, find the coefficient of \(x^4\) in \((3 + x)^5\). Using the binomial theorem, the general term is \(\binom{5}{r} 3^{5-r} x^r\). For \(x^4\), set \(r = 4\):

\(\binom{5}{4} 3^{1} x^4 = 5 \times 3 = 15\).

Next, find the coefficient of \(x^2\) in \(\left(2x + \frac{a}{x}\right)^6\). The general term is \(\binom{6}{r} (2x)^{6-r} \left(\frac{a}{x}\right)^r\).

For \(x^2\), solve \(6-r-r = 2\), giving \(r = 2\).

The term is \(\binom{6}{2} (2x)^{4} \left(\frac{a}{x}\right)^2 = 15 \times 16x^4 \times \frac{a^2}{x^2} = 240a^2 x^2\).

Equating the coefficients: \(15 = 240a^2\).

Solving for \(a\):

\(a^2 = \frac{15}{240} = \frac{1}{16}\).

\(a = \frac{1}{4}\) (since \(a\) is positive).

(a) To find \(a\), consider the term in \(x^4\) from \(\left( 2x^2 + \frac{k^2}{x} \right)^5\). The term is given by:

\([10 \times x](2x^2)^3 \left( \frac{k^2}{x} \right)^2\)

\(80k^4 x^4 \Rightarrow a = 80k^4\).

To find \(b\), consider the term in \(x^2\) from \((2kx - 1)^4\). The term is given by:

\([6 \times 1](2kx)^2 \times 1 = 24k^2 x^2 \Rightarrow b = 24k^2\)

(b) Given \(a + b = 216\), substitute the expressions for \(a\) and \(b\):

\(80k^4 + 24k^2 = 216\)

\(10k^4 + 3k^2 - 27 = 0\)

Factorizing gives:

\((2k^2 - 3)(5k^2 + 9) = 0\)

\(k^2 = \frac{3}{2}\) or \(k^2 = -\frac{9}{5}\)

Since \(k^2\) must be non-negative, \(k^2 = \frac{3}{2}\).

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