(a) To find the term independent of x in \(\left( 3x + \frac{2}{x^2} \right)^6\), use the binomial expansion:

\(\binom{6}{k} (3x)^{6-k} \left( \frac{2}{x^2} \right)^k\).

We need the power of x to be zero:

\(6-k - 2k = 0 \Rightarrow 6 = 3k \Rightarrow k = 2\).

Substitute \(k = 2\):

\(\binom{6}{2} (3x)^4 \left( \frac{2}{x^2} \right)^2 = 15 \times 81x^4 \times \frac{4}{x^4} = 4860\).

(b) For \(\left( 3x + \frac{2}{x^2} \right)^6 (1 - x^3)\), find the term independent of x:

Use the term from part (a) and subtract the term where \(x^3\) is multiplied by the term in \(\left( 3x + \frac{2}{x^2} \right)^6\) that results in \(x^{-3}\).

Find the term in \(\left( 3x + \frac{2}{x^2} \right)^6\) with \(x^{-3}\):

\(6-k - 2k = -3 \Rightarrow 6 = 3k - 3 \Rightarrow k = 3\).

Substitute \(k = 3\):

\(\binom{6}{3} (3x)^3 \left( \frac{2}{x^2} \right)^3 = 20 \times 27x^3 \times \frac{8}{x^6} = 4320\).

Subtract this from the term in part (a):

\(4860 - 4320 = 540\).

To find the term independent of x in the expansion of \(\left( 2x + \frac{1}{4x^2} \right)^6\), we use the binomial theorem:

\(\binom{6}{k} (2x)^{6-k} \left( \frac{1}{4x^2} \right)^k\)

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

\((6-k) - 2k = 0\)

Solving for k gives:

\(6 - 3k = 0\)

\(k = 2\)

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

\(\binom{6}{2} (2x)^4 \left( \frac{1}{4x^2} \right)^2\)

\(= 15 \times 16x^4 \times \frac{1}{16x^4}\)

\(= 15\)

Thus, the term independent of x is 15.

To find the term independent of x, we use the binomial expansion formula:

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

Here, \(a = 2x\) and \(b = -\frac{1}{4x^2}\), and \(n = 9\).

The general term is \(T_k = \binom{9}{k} (2x)^{9-k} \left(-\frac{1}{4x^2}\right)^k\).

Simplifying, \(T_k = \binom{9}{k} \cdot 2^{9-k} \cdot x^{9-k} \cdot (-1)^k \cdot \frac{1}{4^k} \cdot x^{-2k}\).

\(T_k = \binom{9}{k} \cdot 2^{9-k} \cdot (-1)^k \cdot \frac{1}{4^k} \cdot x^{9-k-2k}\).

\(T_k = \binom{9}{k} \cdot 2^{9-k} \cdot (-1)^k \cdot \frac{1}{4^k} \cdot x^{9-3k}\).

For the term to be independent of x, the power of x must be zero:

\(9 - 3k = 0\).

Solving for \(k\), we get \(k = 3\).

Substitute \(k = 3\) into the expression for \(T_k\):

\(T_3 = \binom{9}{3} \cdot 2^{9-3} \cdot (-1)^3 \cdot \frac{1}{4^3}\).

\(T_3 = \binom{9}{3} \cdot 2^6 \cdot (-1) \cdot \frac{1}{64}\).

\(T_3 = 84 \cdot 64 \cdot (-1) \cdot \frac{1}{64}\).

\(T_3 = -84\).

Thus, the term independent of x is \(-84\).

To find the term independent of x, we use the binomial expansion formula:

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

Here, \(a = 2x\) and \(b = \frac{1}{2x^3}\), and \(n = 8\).

We need the term where the power of x is zero. The general term is:

\(T_k = \binom{8}{k} (2x)^{8-k} \left( \frac{1}{2x^3} \right)^k\).

Simplifying, we get:

\(T_k = \binom{8}{k} \cdot 2^{8-k} \cdot x^{8-k} \cdot \frac{1}{2^k} \cdot x^{-3k}\).

\(T_k = \binom{8}{k} \cdot 2^{8-2k} \cdot x^{8-4k}\).

We want the power of x to be zero:

\(8 - 4k = 0\).

Solving for \(k\), we get \(k = 2\).

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

\(T_2 = \binom{8}{2} \cdot 2^{8-4} \cdot x^{0}\).

\(T_2 = 28 \cdot 2^4 \cdot 1\).

\(T_2 = 28 \cdot 16\).

\(T_2 = 448\).

Thus, the term independent of x is 448.

To find the term independent of x, we use the binomial expansion formula:

\(\left( x - \frac{3}{2x} \right)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} \left( -\frac{3}{2x} \right)^k\)

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

\(6-k-k = 0\)

Solving for k, we get:

\(6 - 2k = 0\)

\(k = 3\)

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

\(\binom{6}{3} x^{6-3} \left( -\frac{3}{2x} \right)^3\)

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

\(= 20 \cdot x^3 \cdot \left( -\frac{3}{2x} \right)^3\)

\(= 20 \cdot x^3 \cdot \left( -\frac{27}{8x^3} \right)\)

\(= 20 \cdot \left( -\frac{27}{8} \right)\)

\(= -67.5\)

To find the term independent of x, we need the power of x to be zero in the expansion of \(\left( 4x^3 + \frac{1}{2x} \right)^8\).

Consider the general term in the binomial expansion:

\(T_r = \binom{8}{r} (4x^3)^{8-r} \left( \frac{1}{2x} \right)^r\).

Simplifying, we get:

\(T_r = \binom{8}{r} \cdot 4^{8-r} \cdot x^{3(8-r)} \cdot \frac{1}{2^r} \cdot x^{-r}\).

Combine the powers of x:

\(x^{3(8-r) - r} = x^{24 - 4r}\).

We want the power of x to be zero:

\(24 - 4r = 0\).

Solving for r gives:

\(r = 6\).

Substitute \(r = 6\) back into the expression for the term:

\(T_6 = \binom{8}{6} \cdot 4^2 \cdot \frac{1}{2^6}\).

Calculate \(\binom{8}{6} = 28\), \(4^2 = 16\), and \(2^6 = 64\).

Thus, \(T_6 = 28 \cdot 16 \cdot \frac{1}{64} = 28\).

Therefore, the term independent of x is 28.

To find the term independent of x, we use the binomial expansion formula:

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

Here, \(a = 2x\) and \(b = \frac{1}{x^2}\), and \(n = 6\).

The general term is:

\(T_k = \binom{6}{k} (2x)^{6-k} \left( \frac{1}{x^2} \right)^k\).

Simplifying, we get:

\(T_k = \binom{6}{k} \cdot 2^{6-k} \cdot x^{6-k} \cdot x^{-2k}\).

For the term to be independent of x, the power of x must be zero:

\(6-k - 2k = 0\).

Solving for \(k\), we get:

\(6 - 3k = 0\)

\(3k = 6\)

\(k = 2\).

Substitute \(k = 2\) back into the expression for \(T_k\):

\(T_2 = \binom{6}{2} \cdot 2^{4} \cdot x^{0}\).

\(T_2 = 15 \cdot 16 \cdot 1\).

\(T_2 = 240\).

Thus, the term independent of x is 240.

To find the term independent of x in the expansion of \(\left( x - \frac{1}{x^2} \right)^9\), we use the binomial theorem:

\(\left( x - \frac{1}{x^2} \right)^9 = \sum_{k=0}^{9} \binom{9}{k} x^{9-k} \left( -\frac{1}{x^2} \right)^k\)

The general term is:

\(T_k = \binom{9}{k} x^{9-k} \left( -\frac{1}{x^2} \right)^k = \binom{9}{k} (-1)^k x^{9-k-2k}\)

\(= \binom{9}{k} (-1)^k x^{9-3k}\)

We need the exponent of x to be zero for the term to be independent of x:

\(9 - 3k = 0\)

\(3k = 9\)

\(k = 3\)

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

\(T_3 = \binom{9}{3} (-1)^3 x^{9-3 \times 3}\)

\(= \binom{9}{3} (-1)^3 x^0\)

\(= \binom{9}{3} (-1)^3\)

\(= 84 \times (-1)\)

\(= -84\)

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

\(\left( x + \frac{3}{x} \right)^4 = \sum_{r=0}^{4} \binom{4}{r} x^{4-r} \left( \frac{3}{x} \right)^r\)

The general term is:

\(T_r = \binom{4}{r} x^{4-r} \left( \frac{3}{x} \right)^r = \binom{4}{r} x^{4-r} \cdot \frac{3^r}{x^r} = \binom{4}{r} \cdot 3^r \cdot x^{4-2r}\)

We need the exponent of x to be zero for the term to be independent of x:

\(4 - 2r = 0\)

Solving for r gives:

\(4 = 2r\)

\(r = 2\)

Substitute r back into the general term:

\(T_2 = \binom{4}{2} \cdot 3^2 \cdot x^{4-4} = \binom{4}{2} \cdot 9\)

\(\binom{4}{2} = 6\)

\(T_2 = 6 \cdot 9 = 54\)

Thus, the value of the term independent of x is 54.

(a) To expand \((1 + a)^5\), use the binomial theorem:

\((1 + a)^5 = \sum_{k=0}^{5} \binom{5}{k} a^k\).

Calculate the terms up to \(a^3\):

\(\binom{5}{0} a^0 = 1\)

\(\binom{5}{1} a^1 = 5a\)

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

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

Thus, \((1 + a)^5 = 1 + 5a + 10a^2 + 10a^3 + \ldots\)

(b) To expand \([1 + (x + x^2)]^5\), substitute \(a = x + x^2\) into the expansion from part (a):

\(1 + 5(x + x^2) + 10(x + x^2)^2 + 10(x + x^2)^3 + \ldots\)

Calculate each term up to \(x^3\):

\(5(x + x^2) = 5x + 5x^2\)

\(10(x + x^2)^2 = 10(x^2 + 2x^3 + \ldots) = 10x^2 + 20x^3 + \ldots\)

\(10(x + x^2)^3 = 10(x^3 + \ldots) = 10x^3 + \ldots\)

Combine terms: \(1 + 5x + (5x^2 + 10x^2) + (20x^3 + 10x^3) + \ldots\)

\(= 1 + 5x + 15x^2 + 30x^3 + \ldots\)

(i) To find the first three terms in the expansion of \((2+u)^5\), we use the binomial theorem:

\((2+u)^5 = \sum_{k=0}^{5} \binom{5}{k} 2^{5-k} u^k\).

The first three terms are:

\(\binom{5}{0} 2^5 u^0 = 32\),

\(\binom{5}{1} 2^4 u^1 = 80u\),

\(\binom{5}{2} 2^3 u^2 = 80u^2\).

Thus, the first three terms are \(32 + 80u + 80u^2\).

(ii) Substitute \(u = x + x^2\) into the expansion:

\(80(x + x^2) + 80(x + x^2)^2\).

Calculate the coefficient of \(x^2\):

From \(80(x + x^2)\), the coefficient of \(x^2\) is \(80\).

From \(80(x + x^2)^2\), expand \((x + x^2)^2 = x^2 + 2x^3 + x^4\), the coefficient of \(x^2\) is \(80\).

Adding these gives the total coefficient of \(x^2\) as \(80 + 80 = 160\).

(i) To find the term independent of \(x\) in the expansion of \(\left( \frac{2}{x} - 3x \right)^6\), we use the binomial theorem:

\(\binom{6}{3} \left( \frac{2}{x} \right)^3 (-3x)^3\)

Calculating each part:

\(\binom{6}{3} = 20\)

\(\left( \frac{2}{x} \right)^3 = \frac{8}{x^3}\)

\((-3x)^3 = -27x^3\)

Multiplying these gives:

\(20 \times \frac{8}{x^3} \times (-27x^3) = -4320\)

Thus, the term independent of \(x\) is \(-4320\).

(ii) To find the value of \(a\) for which there is no term independent of \(x\) in the expansion of \(\left( 1 + ax^2 \right) \left( \frac{2}{x} - 3x \right)^6\), we consider the critical term:

\(\binom{6}{2} \left( \frac{2}{x} \right)^4 (-3x)^2\)

Calculating each part:

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

\(\left( \frac{2}{x} \right)^4 = \frac{16}{x^4}\)

\((-3x)^2 = 9x^2\)

Multiplying these gives:

\(15 \times \frac{16}{x^4} \times 9x^2 = \frac{2160}{x^2}\)

For no term independent of \(x\), the coefficient of \(x^0\) must be zero:

\(15a \times 16 \times 9 - 4320 = 0\)

Solving for \(a\):

\(2160a = 4320\)

\(a = 2\)

(i) The general term in the expansion of \(\left( x - \frac{2}{x} \right)^6\) is given by \(T_r = \binom{6}{r} x^{6-r} \left( -\frac{2}{x} \right)^r\).

For the term to be independent of \(x\), the powers of \(x\) must cancel out, i.e., \(6 - r - r = 0\), which gives \(r = 3\).

Substituting \(r = 3\) into the general term, we get:

\(T_3 = \binom{6}{3} x^{6-3} \left( -\frac{2}{x} \right)^3 = 20 \cdot x^3 \cdot \left( -\frac{8}{x^3} \right) = -160\).

Thus, the term independent of \(x\) is \(-160\).

(ii) The term independent of \(x\) in the product \(\left( 2 + \frac{3}{x^2} \right) \left( x - \frac{2}{x} \right)^6\) is found by considering the contributions from both terms.

First, find the term in \(\left( x - \frac{2}{x} \right)^6\) that results in \(x^2\):

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

For \(x^2\), \(6 - r - r = 2\), which gives \(r = 2\).

Substituting \(r = 2\), we get:

\(T_2 = \binom{6}{2} x^{6-2} \left( -\frac{2}{x} \right)^2 = 15 \cdot x^4 \cdot \frac{4}{x^2} = 60x^2\).

Now, consider the product:

\(2 \cdot (-160) + \frac{3}{x^2} \cdot 60x^2 = -320 + 180 = -140\).

Thus, the term independent of \(x\) is \(-140\).

(i) The binomial expansion of \((1 - 2x)^5\) is given by:

\((1 - 2x)^5 = \sum_{k=0}^{5} \binom{5}{k} (1)^{5-k} (-2x)^k\)

For \(x^4\), \(k = 4\):

Coefficient = \(\binom{5}{4} (1)^{1} (-2)^4 = 5 \times 16 = 80\)

For \(x^5\), \(k = 5\):

Coefficient = \(\binom{5}{5} (1)^{0} (-2)^5 = 1 \times (-32) = -32\)

(ii) In the expansion of \((1 + px)(1 - 2x)^5\), the term in \(x^5\) is:

\((-32) + 5p(1)(-2)^4 = 0\)

\(-32 + 80p = 0\)

\(80p = 32\)

\(p = \frac{32}{80} = \frac{2}{5}\)

First, expand \((a + x)^5\) using the binomial theorem:

\((a + x)^5 = a^5 + 5a^4x + 10a^3x^2 + \ldots\)

We are interested in the terms that will contribute to \(x^2\) in the product \(\left( 1 - \frac{2x}{a} \right)(a + x)^5\).

Consider the term \(-\frac{2x}{a} \cdot a^5\) which contributes \(-\frac{2a^5x}{a} = -2a^4x\).

Now consider the term \(1 \cdot 10a^3x^2 = 10a^3x^2\).

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

\(-\frac{2}{a} \times 5a^4 + 10a^3 = -10a^3 + 10a^3 = 0\).

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

(i) To expand \((2-y)^5\), use the binomial theorem:

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

The first three terms are:

\(\binom{5}{0} (2)^5 (-y)^0 = 32\)

\(\binom{5}{1} (2)^4 (-y)^1 = -80y\)

\(\binom{5}{2} (2)^3 (-y)^2 = 80y^2\)

Thus, the first three terms are \(32 - 80y + 80y^2\).

(ii) Substitute \(y = 2x + x^2\) into the expansion:

\((2-(2x-x^2))^5 = (2-(2x+x^2))^5 = (2-y)^5\).

Using the result from part (i), the terms involving \(x^2\) are:

\(-80(2x) + 80(x^2)\).

Calculate the coefficient of \(x^2\):

\(-80 \cdot 2x + 80x^2 = -160x + 80x^2\).

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

\(80 \cdot 4 = 320\).

Thus, the coefficient of \(x^2\) is \(400\).

(i) To find the term in \(x^2\) in the expansion of \((1 - \frac{3}{2}x)^6\), use the binomial theorem:

\(^6C_2 \left( -\frac{3}{2}x \right)^2 = 15 \times \left( \frac{9}{4}x^2 \right) = 135x^2.\)

For the term in \(x^3\):

\(^6C_3 \left( -\frac{3}{2}x \right)^3 = 20 \times \left( -\frac{27}{8}x^3 \right) = -540x^3.\)

(ii) For the expansion of \((k + 2x)(1 - \frac{3}{2}x)^6\), the term in \(x^3\) is:

\(2 \times 135x^2 + k \times (-540x^3) = 0.\)

\(\frac{270x^3}{4} - \frac{135kx^3}{2} = 0.\)

Solving for \(k\):

\(270 - 270k = 0\)

\(k = 1.\)

The general term in the binomial expansion of \((1 + ax)^6\) is given by:

\(T_k = \binom{6}{k} (1)^{6-k} (ax)^k = \binom{6}{k} a^k x^k\)

For the term in \(x\), we have \(k = 1\):

\(T_1 = \binom{6}{1} a^1 x^1 = 6ax\)

We know the coefficient of \(x\) is \(-30\), so:

\(6a = -30\)

Solving for \(a\):

\(a = -5\)

Now, find the coefficient of \(x^3\) (\(k = 3\)):

\(T_3 = \binom{6}{3} a^3 x^3\)

\(\binom{6}{3} = \frac{6 \times 5 \times 4}{3!} = 20\)

Substitute \(a = -5\):

\(T_3 = 20(-5)^3 x^3\)

\(T_3 = 20(-125)x^3\)

\(T_3 = -2500x^3\)

Thus, the coefficient of \(x^3\) is \(-2500\).

(i) The binomial expansion of \((k + x)^8\) is given by:

\(\sum_{r=0}^{8} \binom{8}{r} k^{8-r} x^r\)

The first four terms are:

\(\binom{8}{0} k^8 x^0 = k^8\)

\(\binom{8}{1} k^7 x^1 = 8k^7x\)

\(\binom{8}{2} k^6 x^2 = 28k^6x^2\)

\(\binom{8}{3} k^5 x^3 = 56k^5x^3\)

Thus, the first four terms are \(k^8 + 8k^7x + 28k^6x^2 + 56k^5x^3\).

(ii) Given that the coefficients of \(x^2\) and \(x^3\) are equal, we have:

\(28k^6 = 56k^5\)

Dividing both sides by \(28k^5\), we get:

\(k = 2\)