(i) Express \(\frac{2 - x + 8x^2}{(1-x)(1+2x)(2+x)}\) in the form \(\frac{A}{1-x} + \frac{B}{1+2x} + \frac{C}{2+x}\).

Equating coefficients, solve for \(A\), \(B\), and \(C\):

\(A = 1\), \(B = 2\), \(C = -4\).

Thus, \(\frac{2 - x + 8x^2}{(1-x)(1+2x)(2+x)} = \frac{1}{1-x} + \frac{2}{1+2x} - \frac{4}{2+x}\).

(ii) Expand each partial fraction:

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

\(\frac{2}{1+2x} = 2(1 - 2x + 4x^2 + \ldots) = 2 - 4x + 8x^2 + \ldots\)

\(\frac{-4}{2+x} = -4(\frac{1}{2} - \frac{x}{4} + \frac{x^2}{8} + \ldots) = -2 + x - \frac{x^2}{2} + \ldots\)

Combine the expansions up to \(x^2\):

\(1 + x + x^2 + 2 - 4x + 8x^2 - 2 + x - \frac{x^2}{2} = 1 - 2x + \frac{17}{2}x^2\).

(i) To express \(\frac{10}{(2-x)(1+x^2)}\) in partial fractions, assume \(\frac{10}{(2-x)(1+x^2)} = \frac{A}{2-x} + \frac{Bx+C}{1+x^2}\).

Multiply through by \((2-x)(1+x^2)\) to get:

\(10 = A(1+x^2) + (Bx+C)(2-x)\).

Expanding and equating coefficients, solve for \(A\), \(B\), and \(C\):

\(A = 2\), \(B = 2\), \(C = 4\).

Thus, \(\frac{10}{(2-x)(1+x^2)} = \frac{2}{2-x} + \frac{2x+4}{1+x^2}\).

(ii) For the expansion, consider \(\frac{2}{2-x} = \frac{1}{1-\frac{x}{2}}\) and expand using the geometric series:

\(\frac{2}{2-x} = 1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \ldots\)

For \(\frac{2x+4}{1+x^2}\), expand \(\frac{1}{1+x^2} = 1 - x^2 + x^4 - \ldots\) and multiply by \(2x+4\):

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

Combine the expansions:

\(5 + \frac{5}{2}x - \frac{15}{4}x^2 - \frac{15}{8}x^3\).

(i) Express \(\frac{3x^2 + x}{(x+2)(x^2+1)}\) in partial fractions:

Assume \(\frac{3x^2 + x}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+1}\).

Multiply through by \((x+2)(x^2+1)\) to clear the denominators:

\(3x^2 + x = A(x^2+1) + (Bx+C)(x+2)\).

Expand and equate coefficients:

\(3x^2 + x = Ax^2 + A + Bx^2 + 2Bx + Cx + 2C\).

Combine like terms:

\((A+B)x^2 + (2B+C)x + (A+2C) = 3x^2 + x\).

Equate coefficients:

\(A + B = 3\), \(2B + C = 1\), \(A + 2C = 0\).

Solve the system of equations:

From \(A + 2C = 0\), we have \(A = -2C\).

Substitute \(A = -2C\) into \(A + B = 3\):

\(-2C + B = 3\).

Substitute \(A = -2C\) into \(2B + C = 1\):

\(2B + C = 1\).

Now solve:

From \(-2C + B = 3\), \(B = 3 + 2C\).

Substitute \(B = 3 + 2C\) into \(2B + C = 1\):

\(2(3 + 2C) + C = 1\).

\(6 + 4C + C = 1\).

\(5C = -5\).

\(C = -1\).

\(A = -2(-1) = 2\).

\(B = 3 + 2(-1) = 1\).

Thus, \(\frac{3x^2 + x}{(x+2)(x^2+1)} = \frac{2}{x+2} + \frac{x-1}{x^2+1}\).

(ii) Expand \(\frac{3x^2 + x}{(x+2)(x^2+1)}\) in ascending powers of \(x\):

Use the partial fractions: \(\frac{2}{x+2} + \frac{x-1}{x^2+1}\).

Expand \(\frac{2}{x+2}\) using binomial series:

\(\frac{2}{x+2} = 2(1 + \frac{x}{2})^{-1} \approx 2(1 - \frac{x}{2} + \frac{x^2}{4} - \frac{x^3}{8})\).

Expand \(\frac{x-1}{x^2+1}\):

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

\(= x - 1 - x^3 + x \approx x - 1\).

Add the expansions:

\(2(1 - \frac{x}{2} + \frac{x^2}{4} - \frac{x^3}{8}) + (x - 1)\).

\(= 2 - x + \frac{x^2}{2} - \frac{x^3}{4} + x - 1\).

\(= 1 + \frac{x^2}{2} - \frac{x^3}{4}\).

Combine terms:

\(= \frac{1}{2}x + \frac{5}{4}x^2 - \frac{9}{8}x^3\).

(i) Express \(f(x)\) as partial fractions:

Assume \(f(x) = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x+1}\).

Multiply through by \((x-1)(x-2)(x+1)\) to clear the denominators:

\(x^2 + 7x - 6 = A(x-2)(x+1) + B(x-1)(x+1) + C(x-1)(x-2)\).

Expand and equate coefficients to solve for \(A, B,\) and \(C\):

\(A = -1, B = 4, C = -2\).

Thus, \(f(x) = \frac{-1}{x-1} + \frac{4}{x-2} - \frac{2}{x+1}\).

(ii) Expand each term for small \(x\):

\(\frac{-1}{x-1} = -1 - x - x^2 - x^3 - \ldots\)

\(\frac{4}{x-2} = -2 - 4x - 8x^2 - 16x^3 - \ldots\)

\(\frac{-2}{x+1} = -2 + 2x - 2x^2 + 2x^3 - \ldots\)

Add these expansions:

\(f(x) = (-1 - x - x^2 - x^3) + (-2 - 4x - 8x^2 - 16x^3) + (-2 + 2x - 2x^2 + 2x^3)\)

Combine like terms:

\(f(x) = -3 + 2x - \frac{3}{2}x^2 + \frac{11}{4}x^3\).

(i) Express \(f(x)\) as partial fractions:

Assume \(f(x) = \frac{A}{2x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}\).

Equating coefficients, we find:

\(A = 1\), \(B = 0\), \(C = 4\).

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

(ii) For small \(x\), expand \((1 + 2x)^{-1}\) and \((x-2)^{-2}\):

\((1 + 2x)^{-1} \approx 1 - 2x\)

\((x-2)^{-2} \approx 1 + 4x + 8x^2\)

Multiply these expansions:

\((1 - 2x)(1 + 4x + 8x^2) \approx 1 + 2x + 8x^2\)

Combine terms to get:

\(f(x) \approx 1 - x + 5x^2\).