(i) To find the coefficients of \(x^2\) and \(x^3\) in \((2-x)^6\), use the binomial theorem:

\((2-x)^6 = \sum_{k=0}^{6} \binom{6}{k} (2)^{6-k} (-x)^k\).

The coefficient of \(x^2\) is given by \(\binom{6}{2} (2)^4 (-1)^2 = 15 \times 16 = 240\).

The coefficient of \(x^3\) is given by \(\binom{6}{3} (2)^3 (-1)^3 = 20 \times 8 \times (-1) = -160\).

(ii) For \((3x+1)(2-x)^6\), we need the coefficient of \(x^3\).

Consider the terms that contribute to \(x^3\):

- From \(3x \times \text{coefficient of } x^2 \text{ in } (2-x)^6\): \(3 \times 240 = 720\).

- From \(1 \times \text{coefficient of } x^3 \text{ in } (2-x)^6\): \(-160\).

Thus, the coefficient of \(x^3\) is \(720 - 160 = 560\).

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

The general term in the expansion is given by:

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

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

\(2(5-k) - k = 10 - 3k\)

Setting this equal to 1 gives:

\(10 - 3k = 1\)

\(3k = 9\)

\(k = 3\)

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

\(T_3 = \binom{5}{3} (x^2)^2 \left(-\frac{2}{x}\right)^3\)

\(= 10 \cdot x^4 \cdot \left(-\frac{8}{x^3}\right)\)

\(= 10 \cdot x^4 \cdot \left(-\frac{8}{x^3}\right)\)

\(= 10 \cdot (-8) \cdot x\)

\(= -80x\)

Thus, the coefficient of x is -80.

To find the coefficient of \(x^2\) in the expansion of \((1 + x^2) \left( \frac{x}{2} - \frac{4}{x} \right)^6\), we first expand \(\left( \frac{x}{2} - \frac{4}{x} \right)^6\) using the binomial theorem.

The general term in the expansion is given by:

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

Simplifying, we have:

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

We need the term where the power of \(x\) is zero (constant term) and the term where the power of \(x\) is 2.

For the constant term \(x^0\):

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

Substitute \(k = 3\):

\(\binom{6}{3} \frac{(-4)^3}{2^3} = 20 \times \frac{-64}{8} = -160\)

For the term in \(x^2\):

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

Substitute \(k = 2\):

\(\binom{6}{2} \frac{(-4)^2}{2^4} = 15 \times \frac{16}{16} = 15\)

Now consider the expansion of \((1 + x^2)\):

\(1 + x^2\)

The coefficient of \(x^2\) in the product is:

\(1 \times 15 + x^2 \times (-160) = 15 - 160 = -145\)

Thus, the coefficient of \(x^2\) is \(-145\).

Answer: (i) \(81\). (ii) \(270\). (iii) \(675\).

(i) Write

\((x+3x^2)^4=x^4(1+3x)^4.\)

To obtain the term in \(x^8\), we need the term in \(x^4\) from \((1+3x)^4\), which is

\((3x)^4=81x^4.\)

So the coefficient of \(x^8\) is \(81\).

(ii) Similarly,

\((x+3x^2)^5=x^5(1+3x)^5.\)

To obtain \(x^8\), we need the term in \(x^3\) from \((1+3x)^5\):

\(\binom{5}{3}(3x)^3=10\cdot27x^3=270x^3.\)

So the coefficient of \(x^8\) is \(270\).

(iii) Let \(u=x+3x^2\). Then

\((1+u)^5=1+5u+10u^2+10u^3+5u^4+u^5.\)

Only \(5u^4\) and \(u^5\) can produce an \(x^8\) term.

From part (i), the coefficient of \(x^8\) in \(u^4\) is \(81\), so the contribution from \(5u^4\) is \(5\cdot81=405\).

From part (ii), the coefficient of \(x^8\) in \(u^5\) is \(270\).

Therefore the required coefficient is

\(405+270=675.\)

(i) To find the coefficient of \(x^2\) in \(\left( 2x - \frac{1}{2x} \right)^6\), we use the binomial expansion:

The general term is \(\binom{6}{r} (2x)^{6-r} \left(-\frac{1}{2x}\right)^r\).

We need \((6-r) - r = 2\), which gives \(6 - 2r = 2\), so \(r = 2\).

The term is \(\binom{6}{2} (2x)^4 \left(-\frac{1}{2x}\right)^2\).

Calculating, \(\binom{6}{2} = 15\), \((2x)^4 = 16x^4\), and \(\left(-\frac{1}{2x}\right)^2 = \frac{1}{4x^2}\).

The coefficient of \(x^2\) is \(15 \times 16 \times \frac{1}{4} = 60\).

(ii) For \((1 + x^2) \left( 2x - \frac{1}{2x} \right)^6\), we consider two terms:

1. The constant term from \(\left( 2x - \frac{1}{2x} \right)^6\) is \(20 \times x^3 \times \left(-\frac{1}{2x}\right)^3 = 20 \times 8x^3 \times \left(-\frac{1}{8x^3}\right) = 20\).

2. The \(x^2\) term from \((1 + x^2)\) is \(60\) from part (i).

The coefficient of \(x^2\) is \(60 - 20 = 40\).

To find the coefficient of \(x^3\) in the expansion of \(\left( 2 - \frac{1}{2}x \right)^7\), we use the binomial theorem:

\(\binom{n}{k} a^{n-k} b^k\)

where \(n = 7\), \(a = 2\), \(b = -\frac{1}{2}x\), and we need the term where the power of \(x\) is 3.

Thus, \(k = 3\) and the term is:

\(\binom{7}{3} (2)^{7-3} \left(-\frac{1}{2}x\right)^3\)

Calculate each part:

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

\((2)^4 = 16\)

\(\left(-\frac{1}{2}\right)^3 = -\frac{1}{8}\)

Combine these to find the coefficient:

\(35 \times 16 \times -\frac{1}{8} = -70\)

Therefore, the coefficient of \(x^3\) is \(-70\).

To find the coefficient of \(x^6\) in the expansion of \(\left( 2x^3 - \frac{1}{x^2} \right)^7\), we use the binomial theorem:

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

Here, \(a = 2x^3\) and \(b = -\frac{1}{x^2}\), and \(n = 7\).

We need the term where the power of \(x\) is 6. The general term is:

\(T_k = \binom{7}{k} (2x^3)^{7-k} \left(-\frac{1}{x^2}\right)^k\)

Simplifying, we get:

\(T_k = \binom{7}{k} \cdot 2^{7-k} \cdot x^{3(7-k)} \cdot (-1)^k \cdot x^{-2k}\)

\(= \binom{7}{k} \cdot 2^{7-k} \cdot (-1)^k \cdot x^{21-3k-2k}\)

\(= \binom{7}{k} \cdot 2^{7-k} \cdot (-1)^k \cdot x^{21-5k}\)

We want \(21-5k = 6\), so:

\(21 - 5k = 6\)

\(5k = 15\)

\(k = 3\)

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

\(T_3 = \binom{7}{3} \cdot 2^{4} \cdot (-1)^3 \cdot x^{6}\)

\(= 35 \cdot 16 \cdot (-1) \cdot x^{6}\)

\(= -560 \cdot x^{6}\)

Thus, the coefficient of \(x^6\) is \(-560\).

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

\(\left( x + \frac{2}{x^2} \right)^7 = \sum_{k=0}^{7} \binom{7}{k} x^{7-k} \left( \frac{2}{x^2} \right)^k\)

We need the term where the power of \(x\) is 1:

\(7-k - 2k = 1\)

Solving for \(k\):

\(7 - 3k = 1\)

\(3k = 6\)

\(k = 2\)

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

\(\binom{7}{2} x^{7-2} \left( \frac{2}{x^2} \right)^2 = \binom{7}{2} x^5 \left( \frac{4}{x^4} \right)\)

\(= \binom{7}{2} \cdot 4 \cdot x = 21 \cdot 4 \cdot x = 84x\)

Thus, the coefficient of \(x\) is 84.

To find the coefficient of \(x^2\) in the expansion of \(\left( \frac{x}{2} + \frac{2}{x} \right)^6\), we use the binomial theorem.

The general term in the expansion is given by:

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

We need the term where the power of \(x\) is 2:

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

\(6 - 2k = 2\)

\(2k = 4\)

\(k = 2\)

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

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

\(= 15 \left( \frac{x^4}{16} \right) \left( \frac{4}{x^2} \right)\)

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

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

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

Thus, the coefficient of \(x^2\) is \(\frac{15}{4}\) or 3.75.

To find the coefficient of \(x^2\) in the expansion of \(\left( x + \frac{2}{x} \right)^6\), we use the binomial theorem.

The general term in the expansion is given by:

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

We need the term where the power of \(x\) is 2:

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

\(6 - 2k = 2\)

\(2k = 4\)

\(k = 2\)

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

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

\(= \binom{6}{2} x^4 \cdot \frac{4}{x^2}\)

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

\(= 15 \cdot 4 \cdot x^2\)

\(= 60x^2\)

Thus, the coefficient of \(x^2\) is 60.

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

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

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

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

The power of \(x\) in this term is \(k - (5-k) = 2k - 5\). We set \(2k - 5 = 1\) to find \(k\):

\(2k - 5 = 1\)

\(2k = 6\)

\(k = 3\)

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

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

\(= 10 \cdot \frac{4}{x^2} \cdot (-27x^3)\)

\(= 10 \cdot 4 \cdot (-27) \cdot x\)

\(= -1080x\)

Thus, the coefficient of \(x\) is \(-1080\).

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

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

Here, \(a = 3x\) and \(b = -\frac{2}{x}\), and \(n = 5\).

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

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

The power of \(x\) in \(T_k\) is given by:

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

We set \(5 - 2k = 1\) to find \(k\):

\(5 - 2k = 1\)

\(2k = 4\)

\(k = 2\)

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

\(T_2 = \binom{5}{2} (3x)^{3} \left(-\frac{2}{x}\right)^2\)

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

\(= 10 \cdot 27 \cdot 4 \cdot x\)

\(= 1080x\)

Thus, the coefficient of \(x\) is 1080.

(i) To find the coefficient of \(x^3\) in \((1 + 2x)^6\), use the binomial theorem:

The general term is \(\binom{6}{r} (1)^{6-r} (2x)^r\).

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

\(\binom{6}{3} (2x)^3 = 20 \times 8x^3 = 160x^3\).

Thus, the coefficient is 160.

(ii) For \((1 - 3x)(1 + 2x)^6\), expand \((1 + 2x)^6\) and find the coefficient of \(x^2\):

The term in \(x^2\) is \(\binom{6}{2} (2x)^2 = 15 \times 4x^2 = 60x^2\).

Now, consider the product \((1 - 3x)(1 + 2x)^6\):

The coefficient of \(x^3\) is given by:

\(1 \times 160 - 3 \times 60 = 160 - 180 = -20\).

Thus, the coefficient is -20.

Answer: \(-40\).

The general term in the expansion of \(\left(2x-\dfrac{1}{x}\right)^5\) is

\(\binom{5}{r}(2x)^{5-r}\left(-\frac{1}{x}\right)^r.\)

The power of \(x\) in this term is

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

We need the coefficient of \(\dfrac{1}{x}\), so

\(5-2r=-1.\)

Hence \(r=3\).

The required term is

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

So

\(\binom{5}{3}(2x)^2\left(-\frac{1}{x}\right)^3=10\cdot4x^2\cdot\left(-\frac{1}{x^3}\right)=-\frac{40}{x}.\)

Therefore the coefficient of \(\dfrac{1}{x}\) is \(-40\).

To find the coefficient of \(\frac{1}{x^3}\) in the expansion of \(\left( x - \frac{2}{x} \right)^7\), we use the binomial theorem:

\(\left( x - \frac{2}{x} \right)^7 = \sum_{k=0}^{7} \binom{7}{k} x^{7-k} \left( -\frac{2}{x} \right)^k\)

We need the term where the power of \(x\) is \(-3\):

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

Set the exponent of \(x\) to \(-3\):

\(7 - 2k = -3\)

\(7 + 3 = 2k\)

\(10 = 2k\)

\(k = 5\)

Substitute \(k = 5\) into the binomial coefficient and expression:

\(\binom{7}{5} x^{7-5} \left( -\frac{2}{x} \right)^5 = \binom{7}{5} x^2 (-2)^5 x^{-5}\)

\(= \binom{7}{5} x^{2-5} (-32)\)

\(= \binom{7}{5} x^{-3} (-32)\)

\(= 21 \cdot (-32) x^{-3}\)

\(= -672 x^{-3}\)

Thus, the coefficient of \(\frac{1}{x^3}\) is \(-672\).

To find the coefficient of \(\frac{1}{x^2}\) in the expansion of \(\left( 3x + \frac{2}{3x^2} \right)^7\), we use the binomial theorem:

\(\left( 3x + \frac{2}{3x^2} \right)^7 = \sum_{k=0}^{7} \binom{7}{k} (3x)^{7-k} \left( \frac{2}{3x^2} \right)^k\)

We need the term where the power of \(x\) is \(-2\):

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

Set the power of \(x\) to \(-2\):

\(7 - 3k = -2\)

\(3k = 9\)

\(k = 3\)

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

\(\binom{7}{3} (3x)^{4} \left( \frac{2}{3x^2} \right)^3\)

\(= \binom{7}{3} \cdot 3^4 \cdot \frac{2^3}{3^3} \cdot x^{4-6}\)

\(= 35 \cdot 81 \cdot \frac{8}{27} \cdot x^{-2}\)

\(= 35 \cdot 3 \cdot 8 \cdot x^{-2}\)

\(= 840 \cdot x^{-2}\)

Thus, the coefficient of \(\frac{1}{x^2}\) is 840.

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

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

We need the term where the power of \(x\) is \(-1\):

\(x^{5-k} \cdot x^{-k} = x^{5-2k}\)

Set \(5-2k = -1\):

\(5 - 2k = -1\)

\(2k = 6\)

\(k = 3\)

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

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

\(= \binom{5}{3} x^2 \left( -\frac{8}{x^3} \right)\)

\(= \binom{5}{3} \cdot (-8) \cdot x^{-1}\)

\(= 10 \cdot (-8) \cdot x^{-1}\)

\(= -80 \cdot x^{-1}\)

Thus, the coefficient of \(\frac{1}{x}\) is \(-80\).

(i) To find the coefficient of \(x^2\) in \((1 - 2x)^7\), use the binomial theorem: \(\binom{7}{2}(-2x)^2 = 21 \times 4x^2 = 84x^2\). The coefficient is 84.

For \(x^3\), \(\binom{7}{3}(-2x)^3 = 35 \times (-8x^3) = -280x^3\). The coefficient is -280.

(ii) The coefficient of \(x^3\) in \((2 + 5x)(1 - 2x)^7\) is found by expanding and combining terms: \(2 \times (-280) + 5 \times 84 = -560 + 420 = -140\).

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

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

We need the term where the power of \(x\) is 1:

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

Set \(5-2k = 1\) to find \(k\):

\(5 - 2k = 1\)

\(2k = 4\)

\(k = 2\)

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

\(\binom{5}{2} (2x)^{3} \left(-\frac{1}{x}\right)^2 = 10 \cdot 8x^3 \cdot \frac{1}{x^2} = 80x\)

Thus, the coefficient of \(x\) is 80.

(ii) Now, find the coefficient of \(x\) in the expansion of \((1 + 3x^2) \left(2x - \frac{1}{x}\right)^5\).

We already know the coefficient of \(x\) in \(\left(2x - \frac{1}{x}\right)^5\) is 80.

We also need the coefficient of \(\frac{1}{x}\) in \(\left(2x - \frac{1}{x}\right)^5\):

Set \(5-2k = -1\):

\(5 - 2k = -1\)

\(2k = 6\)

\(k = 3\)

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

\(\binom{5}{3} (2x)^{2} \left(-\frac{1}{x}\right)^3 = 10 \cdot 4x^2 \cdot \left(-\frac{1}{x^3}\right) = -40 \cdot \frac{1}{x}\)

Now, combine these with \((1 + 3x^2)\):

Coefficient of \(x\) is:

\(1 \times 80 + 3 \times (-40) = 80 - 120 = -40\)

Thus, the coefficient of \(x\) is -40.

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

\(\sum_{k=0}^{5} \binom{5}{k} \left( \frac{1}{x} \right)^{5-k} (3x^2)^k\)

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

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

This simplifies to:

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

We set the exponent of \(x\) to 1:

\(3k - 5 = 1\)

Solving for \(k\):

\(3k = 6\)

\(k = 2\)

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

\(\binom{5}{2} \cdot 3^2 = 10 \cdot 9 = 90\)

Thus, the coefficient of \(x\) is 90.

To find the coefficient of \(x\) in the expansion of \(\left( \frac{x}{3} + \frac{9}{x^2} \right)^7\), we use the binomial theorem.

The general term in the expansion is given by:

\(\binom{7}{r} \left( \frac{x}{3} \right)^{7-r} \left( \frac{9}{x^2} \right)^r\)

Simplifying, we have:

\(\binom{7}{r} \left( \frac{x^{7-r}}{3^{7-r}} \right) \left( \frac{9^r}{x^{2r}} \right) = \binom{7}{r} \frac{9^r}{3^{7-r}} x^{7-r-2r}\)

We need the power of \(x\) to be 1, so:

\(7 - r - 2r = 1\)

\(7 - 3r = 1\)

\(3r = 6\)

\(r = 2\)

Substitute \(r = 2\) back into the expression for the coefficient:

\(\binom{7}{2} \frac{9^2}{3^5}\)

\(= 21 \times \frac{81}{243}\)

\(= 21 \times \frac{1}{3}\)

\(= 7\)

Thus, the coefficient of \(x\) is 7.