(a) Express \(f(x)\) in partial fractions:

Assume \(\frac{17x^2 - 7x + 16}{(2 + 3x^2)(2 - x)} = \frac{Ax + B}{2 + 3x^2} + \frac{C}{2 - x}\).

Multiply through by the denominator: \(17x^2 - 7x + 16 = (Ax + B)(2 - x) + C(2 + 3x^2)\).

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

\(A = -2, B = 3, C = 5\).

Thus, \(f(x) = \frac{-2x + 3}{2 + 3x^2} + \frac{5}{2 - x}\).

(b) Expand \(f(x)\) in ascending powers of \(x\):

Use the binomial expansion for \((2 - x)^{-1}\) and \((2 + 3x^2)^{-1}\):

\((2 - x)^{-1} = \frac{1}{2} + \frac{x}{4} + \frac{x^2}{8} + \frac{x^3}{16}\).

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

Substitute back into the partial fractions and expand:

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

\(\frac{5}{2 - x} = 5\left(\frac{1}{2} + \frac{x}{4} + \frac{x^2}{8} + \frac{x^3}{16}\right)\).

Combine and simplify to get:

\(4 + \frac{1}{4}x - \frac{13}{8}x^2 + \frac{29}{16}x^3\).

(c) The expansion is valid for \(|x| < \sqrt[3]{\frac{2}{3}}\).

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

Assume \(\frac{16 - 17x}{(2 + x)(3 - x)^2} = \frac{A}{2 + x} + \frac{B}{3 - x} + \frac{C}{(3 - x)^2}\).

Multiply through by the denominator \((2 + x)(3 - x)^2\) to clear the fractions:

\(16 - 17x = A(3 - x)^2 + B(2 + x)(3 - x) + C(2 + x)\).

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

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

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Use the partial fraction decomposition:

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

\(\frac{-2}{3 - x} = -\frac{2}{3} - \frac{2x}{9} - \frac{2x^2}{27} + \cdots\)

\(\frac{-1}{(3 - x)^2} = -\frac{1}{9} - \frac{2x}{27} - \frac{3x^2}{81} + \cdots\)

Add these expansions up to the \(x^2\) term:

\(f(x) = \left(1 - \frac{x}{2} + \frac{x^2}{4}\right) + \left(-\frac{2}{3} - \frac{2x}{9} - \frac{2x^2}{27}\right) + \left(-\frac{1}{9} - \frac{2x}{27} - \frac{3x^2}{81}\right)\).

Combine like terms:

\(f(x) = \frac{8}{9} - \frac{43}{54}x + \frac{7}{108}x^2\).

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

Assume \(\frac{12 + 12x - 4x^2}{(2+x)(3-2x)} = \frac{A}{2+x} + \frac{B}{3-2x}\).

Multiply through by \((2+x)(3-2x)\) to clear the denominators:

\(12 + 12x - 4x^2 = A(3-2x) + B(2+x)\).

Expand and collect terms:

\(12 + 12x - 4x^2 = (3A + 2B) + (-2A + B)x\).

Equate coefficients to solve for \(A\) and \(B\):

\(3A + 2B = 12\) and \(-2A + B = 12\).

Solving these equations gives \(A = 2\), \(B = -4\), and \(C = 6\).

Thus, \(f(x) = \frac{2}{2+x} + \frac{-4}{3-2x}\).

(ii) Expansion in ascending powers of \(x\):

Expand \(\frac{1}{2+x}\) using the binomial series:

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

Expand \(\frac{1}{3-2x}\) using the binomial series:

\(\frac{1}{3-2x} = \frac{1}{3}(1 + \frac{2x}{3} + (\frac{2x}{3})^2 + \ldots) = \frac{1}{3} + \frac{2x}{9} + \frac{4x^2}{27} + \ldots\).

Substitute back into the partial fractions:

\(f(x) = 2(\frac{1}{2} - \frac{x}{4} + \frac{x^2}{8}) - 4(\frac{1}{3} + \frac{2x}{9} + \frac{4x^2}{27})\).

Simplify to obtain:

\(f(x) = 2 + \frac{7}{3}x + \frac{7}{18}x^2\).

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

Assume \(\frac{7x^2 - 15x + 8}{(1 - 2x)(2 - x)^2} = \frac{A}{1 - 2x} + \frac{B}{2 - x} + \frac{C}{(2 - x)^2}\).

Equating coefficients, solve:

\(7 = A + 2B\)

\(-15 = -4A - 5B - 2C\)

\(8 = 4A + 2B + C\)

Solving gives \(A = 1\), \(B = 3\), \(C = -2\).

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Expand each term:

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

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

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

Combine terms:

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

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

Combine and simplify:

\(= \frac{9}{4} + x + 4x^2\)

(i) To express \(f(x)\) in the form \(\frac{A}{3-x} + \frac{Bx+C}{2+x^2}\), we equate:

\(\frac{x - 4x^2}{(3-x)(2+x^2)} = \frac{A}{3-x} + \frac{Bx+C}{2+x^2}\)

Multiply through by \((3-x)(2+x^2)\) to clear the denominators:

\(x - 4x^2 = A(2+x^2) + (Bx+C)(3-x)\)

Expand and collect like terms:

\(x - 4x^2 = 2A + Ax^2 + 3Bx + Cx - Bx^2 - Cx\)

\(x - 4x^2 = (A-B)x^2 + (3B+C)x + 2A\)

Equating coefficients, we get:

\(A - B = -4\)

\(3B + C = 1\)

\(2A = 0\) so \(A = 0\)

Substitute \(A = 0\) into \(A - B = -4\):

\(-B = -4\) so \(B = 4\)

Substitute \(B = 4\) into \(3B + C = 1\):

\(12 + C = 1\) so \(C = -11\)

Thus, \(A = 0\), \(B = 4\), \(C = -11\).

(ii) Expand \(\frac{1}{3-x}\) and \(\frac{4x-11}{2+x^2}\) using binomial series:

\(\frac{1}{3-x} = \frac{1}{3} \left(1 + \frac{x}{3} + \frac{x^2}{9} + \frac{x^3}{27} \right)\)

\(\frac{4x-11}{2+x^2} = \frac{4x-11}{2} \left(1 - \frac{x^2}{2} + \frac{x^4}{4} \right)\)

Combine and simplify terms up to \(x^3\):

\(\frac{1}{6}x - \frac{11}{18}x^2 - \frac{31}{108}x^3\)

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

Assume \(f(x) = \frac{A}{x-1} + \frac{B}{3x+2}\).

Multiply through by \((x-1)(3x+2)\) to clear the denominators:

\(12x^2 + 4x - 1 = A(3x+2) + B(x-1)\).

Expand and equate coefficients:

\(12x^2 + 4x - 1 = (3A + B)x + (2A - B)\).

Equating coefficients gives:

\(3A + B = 12\)

\(2A - B = 4\)

Solving these equations, we find \(A = 4\) and \(B = 3\).

Thus, \(f(x) = \frac{4}{x-1} + \frac{3}{3x+2}\).

(ii) Expansion in ascending powers of \(x\):

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

Expand \(\frac{1}{3x+2}\) as \(\frac{1}{2} - \frac{3}{4}x + \frac{9}{8}x^2 - \ldots\)

Combine these expansions:

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

\(= -4 - 4x - 4x^2 + \frac{3}{2} - \frac{9}{4}x + \frac{27}{8}x^2\)

Combine like terms:

\(= \frac{1}{2} - \frac{9}{4}x - \frac{33}{8}x^2\)

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

Assume \(\frac{8x^2 + 9x + 8}{(1-x)(2x+3)^2} = \frac{A}{1-x} + \frac{B}{2x+3} + \frac{C}{(2x+3)^2}\).

Multiply through by \((1-x)(2x+3)^2\) to clear the denominators:

\(8x^2 + 9x + 8 = A(2x+3)^2 + B(1-x)(2x+3) + C(1-x)\).

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

\(A = 1, B = -2, C = 5\).

Thus, \(f(x) = \frac{1}{1-x} - \frac{2}{2x+3} + \frac{5}{(2x+3)^2}\).

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Use the binomial expansion for each term:

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

\(\frac{1}{2x+3} = \frac{1}{3} - \frac{2}{9}x + \cdots\)

\(\frac{1}{(2x+3)^2} = \frac{1}{9} - \frac{4}{27}x + \cdots\)

Combine the expansions:

\(f(x) = \left(1 + x + x^2 + \cdots\right) - 2\left(\frac{1}{3} - \frac{2}{9}x + \cdots\right) + 5\left(\frac{1}{9} - \frac{4}{27}x + \cdots\right)\).

Simplify to obtain the expansion up to \(x^2\):

\(f(x) = \frac{8}{9} + \frac{19}{27}x + \frac{13}{9}x^2\).

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

Assume \(f(x) = \frac{A}{3x+2} + \frac{Bx+C}{x^2+5}\).

Multiply through by the denominator \((3x+2)(x^2+5)\) to clear the fractions:

\(5x^2 - 7x + 4 = A(x^2 + 5) + (Bx + C)(3x + 2)\).

Expand and equate coefficients:

\(5x^2 - 7x + 4 = Ax^2 + 5A + 3Bx^2 + 2Bx + 3Cx + 2C\).

Combine like terms:

\((A + 3B)x^2 + (2B + 3C)x + (5A + 2C)\).

Equate coefficients with \(5x^2 - 7x + 4\):

\(A + 3B = 5\)

\(2B + 3C = -7\)

\(5A + 2C = 4\)

Solve these equations to find \(A = 2\), \(B = 1\), \(C = -3\).

Thus, \(f(x) = \frac{2}{3x+2} + \frac{x-3}{x^2+5}\).

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Expand \(\frac{1}{3x+2}\) using geometric series:

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

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

Expand \(\frac{x-3}{x^2+5}\):

\(\frac{x-3}{x^2+5} = (x-3)(\frac{1}{5} - \frac{1}{5^2}x^2)\)

\(= \frac{1}{5}x - \frac{3}{5} - \frac{1}{25}x^3 + \frac{3}{25}x^2\).

Combine the expansions:

\(f(x) = \frac{2}{5} - \frac{13}{10}x + \frac{237}{100}x^2\).

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

Assume \(\frac{x(6-x)}{(2+x)(4+x^2)} = \frac{A}{2+x} + \frac{Bx+C}{4+x^2}\).

Multiply through by the denominator \((2+x)(4+x^2)\) to clear fractions:

\(x(6-x) = A(4+x^2) + (Bx+C)(2+x)\).

Expand and equate coefficients:

\(x(6-x) = 4A + Ax^2 + 2Bx + Bx^2 + Cx + Cx^2\).

\(x(6-x) = (A+B+C)x^2 + (2B+C)x + 4A\).

Equate coefficients:

\(A + B + C = 0\)

\(2B + C = 6\)

\(4A = 0\)

From \(4A = 0\), \(A = -2\).

Substitute \(A = -2\) into \(A + B + C = 0\):

\(-2 + B + C = 0 \Rightarrow B + C = 2\).

Substitute \(B + C = 2\) into \(2B + C = 6\):

\(2B + C = 6 \Rightarrow 2B + (2-B) = 6 \Rightarrow B = 1\).

Then \(C = 2 - B = 1\).

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

(ii) Expansion in ascending powers of \(x\):

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

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

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

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

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

Combine terms up to \(x^2\):

\(f(x) \approx -2 + x + x + 4 - \frac{x^2}{4}\).

Final answer: \(\frac{3}{4}x - \frac{1}{2}x^2\).

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

Assume \(\frac{3x^2 + x + 6}{(x+2)(x^2+4)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+4}\).

Multiply through by \((x+2)(x^2+4)\) to clear the denominators:

\(3x^2 + x + 6 = A(x^2+4) + (Bx+C)(x+2)\).

Expand and collect like terms:

\(3x^2 + x + 6 = Ax^2 + 4A + Bx^2 + 2Bx + Cx + 2C\).

Combine terms:

\(3x^2 + x + 6 = (A+B)x^2 + (2B+C)x + (4A+2C)\).

Equate coefficients:

\(A + B = 3\)

\(2B + C = 1\)

\(4A + 2C = 6\)

Solve these equations to find \(A = 2\), \(B = 1\), \(C = -1\).

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

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Use the binomial expansion for \((x+2)^{-1}\) and \((x^2+4)^{-1}\):

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

\((x^2+4)^{-1} = \frac{1}{4} - \frac{1}{16}x^2 + \cdots\)

Substitute back into the partial fractions:

\(\frac{2}{x+2} = 2 \left( \frac{1}{2} - \frac{1}{4}x \right) = 1 - \frac{1}{2}x\)

\(\frac{x-1}{x^2+4} = (x-1) \left( \frac{1}{4} - \frac{1}{16}x^2 \right) = \frac{1}{4}x - \frac{1}{4} - \frac{1}{16}x^3 + \cdots\)

Add the expansions:

\(f(x) = 1 - \frac{1}{2}x + \frac{1}{4}x - \frac{1}{4} + \cdots\)

Combine like terms:

\(f(x) = \frac{3}{4} - \frac{1}{4}x + \frac{5}{16}x^2\).

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

Assume \(\frac{10x - 2x^2}{(x+3)(x-1)^2} = \frac{A}{x+3} + \frac{B}{x-1} + \frac{C}{(x-1)^2}\).

Multiply through by \((x+3)(x-1)^2\) to clear the denominators:

\(10x - 2x^2 = A(x-1)^2 + B(x+3)(x-1) + C(x+3)\).

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

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

(ii) Expand in ascending powers of \(x\):

Use the expansions:

\((x+3)^{-1} = \frac{1}{3} - \frac{x}{9} + \frac{x^2}{27} + \ldots\)

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

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

Substitute and simplify:

\(\frac{-3}{x+3} = -\frac{1}{3} + \frac{x}{9} - \frac{x^2}{27} + \ldots\)

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

\(\frac{2}{(x-1)^2} = 2 + 4x + 6x^2 + \ldots\)

Add the series up to \(x^2\):

\(f(x) = \left(-\frac{1}{3} + \frac{x}{9} - \frac{x^2}{27}\right) + (1 + x + x^2) + (2 + 4x + 6x^2)\)

Combine terms:

\(f(x) = \frac{10}{3}x + \frac{44}{9}x^2 + \ldots\)

(a) Express \(f(x)\) in partial fractions:

Assume \(\frac{24x + 13}{(1 - 2x)(2 + x)^2} = \frac{A}{1 - 2x} + \frac{B}{2 + x} + \frac{C}{(2 + x)^2}\).

Multiply through by the denominator \((1 - 2x)(2 + x)^2\) to clear the fractions:

\(24x + 13 = A(2 + x)^2 + B(1 - 2x)(2 + x) + C(1 - 2x)\).

Expand and equate coefficients to find \(A = 4\), \(B = 2\), \(C = -7\).

(b) Expand each partial fraction:

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

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

\(\frac{C}{(2 + x)^2} = -7(2 + x)^{-2} \approx -7(\frac{1}{4} - \frac{x}{4} + \frac{3x^2}{16})\).

Combine these to get:

\(f(x) \approx \frac{13}{4} + \frac{37}{4}x + \frac{239}{16}x^2\).

(c) The expansion is valid for \(|x| < \frac{1}{2}\).

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

Assume \(\frac{4x^2 + 12}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2}\).

Multiply through by \((x+1)(x-3)^2\) to clear the denominators:

\(4x^2 + 12 = A(x-3)^2 + B(x+1)(x-3) + C(x+1)\).

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

\(A = 1, B = 3, C = 12\).

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Use the partial fraction decomposition:

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

\(\frac{1}{x-3} = -\frac{1}{3} - \frac{1}{9}x - \ldots\)

\(\frac{1}{(x-3)^2} = \frac{1}{9} + \frac{2}{27}x + \ldots\)

Combine these expansions:

\(f(x) = \frac{1}{x+1} + 3\left(\frac{1}{x-3}\right) + 12\left(\frac{1}{(x-3)^2}\right)\)

\(= \left(1 - x + x^2\right) + 3\left(-\frac{1}{3} - \frac{1}{9}x\right) + 12\left(\frac{1}{9} + \frac{2}{27}x\right)\)

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

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

Assume \(\frac{5x^2 + x + 6}{(3 - 2x)(x^2 + 4)} = \frac{A}{3 - 2x} + \frac{Bx + C}{x^2 + 4}\).

Multiply through by the denominator \((3 - 2x)(x^2 + 4)\) to get:

\(5x^2 + x + 6 = A(x^2 + 4) + (Bx + C)(3 - 2x)\).

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

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

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

(ii) Expand \(f(x)\) in ascending powers of \(x\):

For \(\frac{3}{3 - 2x}\), use the expansion \((3 - 2x)^{-1} = \frac{1}{3}(1 - \frac{2}{3}x)^{-1}\).

First two terms: \(\frac{1}{3}(1 + \frac{2}{3}x + (\frac{2}{3}x)^2) = \frac{1}{3} + \frac{2}{9}x + \frac{4}{27}x^2\).

For \(\frac{-x - 2}{x^2 + 4}\), use the expansion \((x^2 + 4)^{-1} = \frac{1}{4}(1 - \frac{1}{4}x^2)\).

First two terms: \(\frac{-x - 2}{4}(1 - \frac{1}{4}x^2) = \frac{-x}{4} - \frac{1}{2} + \frac{x^3}{16}\).

Combine terms up to \(x^2\):

\(\frac{1}{3} + \frac{2}{9}x + \frac{4}{27}x^2 - \frac{x}{4} - \frac{1}{2}\).

Simplify to get the final answer:

\(\frac{1}{2} + \frac{5}{12}x + \frac{41}{72}x^2\).

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

Assume \(\frac{x^2 - 8x + 9}{(1-x)(2-x)^2} = \frac{A}{1-x} + \frac{B}{2-x} + \frac{C}{(2-x)^2}\).

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

\(x^2 - 8x + 9 = A(2-x)^2 + B(1-x)(2-x) + C(1-x)\).

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

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

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Use the partial fractions: \(\frac{2}{1-x} - \frac{1}{2-x} + \frac{3}{(2-x)^2}\).

Expand each term using binomial series:

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

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

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

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

\(\frac{9}{4} + \frac{5}{2}x + \frac{39}{16}x^2\).

(i) We express \(\frac{4 + 12x + x^2}{(3-x)(1+2x)^2}\) in partial fractions. Assume the form:

\(\frac{A}{3-x} + \frac{B}{1+2x} + \frac{C}{(1+2x)^2}\)

Equating coefficients, we find:

\(A = 1, \ B = \frac{3}{2}, \ C = -\frac{1}{2}\)

(ii) To expand \(\frac{4 + 12x + x^2}{(3-x)(1+2x)^2}\), we first expand \((3-x)^{-1}\) and \((1+2x)^{-2}\):

\((3-x)^{-1} = \frac{1}{3} + \frac{1}{9}x + \cdots\)

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

Using these expansions, we find the expansion of each partial fraction up to \(x^2\):

\(\frac{1}{3} - \frac{1}{9}x + \frac{1}{27}x^2\)

Combining these, we obtain:

\(\frac{4}{3} - \frac{8}{9}x + \frac{1}{27}x^2\)

(i) To express \(\frac{7x^2 + 8}{(1+x)^2(2-3x)}\) in partial fractions, assume the form:

\(\frac{D}{(1+x)^2} + \frac{E}{1+x} + \frac{F}{2-3x}\)

Equating and solving for constants, we find:

\(D = -1\), \(E = 2\), \(F = 4\).

Thus, the partial fraction decomposition is:

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

(ii) To expand \(\frac{7x^2 + 8}{(1+x)^2(2-3x)}\), use the expansions:

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

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

Combine these expansions to find the first three terms:

\(4 - 2x + \frac{25}{2}x^2\).

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

Assume \(\frac{2x^2 - 7x - 1}{(x-2)(x^2+3)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+3}\).

Multiply through by \((x-2)(x^2+3)\) to clear the denominators:

\(2x^2 - 7x - 1 = A(x^2+3) + (Bx+C)(x-2)\).

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

\(A = -1, B = 3, C = -1\).

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Use the partial fractions: \(\frac{-1}{x-2} + \frac{3x-1}{x^2+3}\).

Expand \(\left(1 - \frac{x}{2}\right)^{-1}\) and \((x^2+3)^{-1}\) using binomial series:

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

\((x^2+3)^{-1} = \frac{1}{3} - \frac{x^2}{9} + \cdots\)

Substitute and multiply out:

\(\frac{1}{6} - \frac{5}{4}x + \frac{17}{72}x^2\).

(i) Express \(\frac{9 - 7x + 8x^2}{(3-x)(1+x^2)}\) in the form \(\frac{A}{3-x} + \frac{Bx+C}{1+x^2}\).

Equating coefficients, we find:

\(A = 6\), \(B = -2\), \(C = 1\).

(ii) Expand \(\frac{1}{3-x}\) as \(\frac{1}{3} \left( 1 + \frac{x}{3} + \frac{x^2}{9} + \frac{x^3}{27} \right)\).

Expand \(\frac{Bx+C}{1+x^2}\) as \((Bx+C)(1-x^2)\).

Combine the expansions to obtain:

\(3 - \frac{4}{3}x - \frac{7}{9}x^2 + \frac{56}{27}x^3\).

(i) Express \(\frac{5x - x^2}{(1+x)(2+x^2)}\) in partial fractions:

Assume \(\frac{5x - x^2}{(1+x)(2+x^2)} = \frac{A}{1+x} + \frac{Bx + C}{2+x^2}\).

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

\(5x - x^2 = A(2+x^2) + (Bx + C)(1+x)\).

Expand and collect like terms:

\(5x - x^2 = 2A + Ax^2 + Bx + Bx^2 + C + Cx\).

Equate coefficients:

For \(x^2\): \(-1 = A + B\)

For \(x\): \(5 = B + C\)

For constant term: \(0 = 2A + C\)

Solve the system of equations:

1. \(A + B = -1\)

2. \(B + C = 5\)

3. \(2A + C = 0\)

From equation 3: \(C = -2A\).

Substitute \(C = -2A\) into equation 2: \(B - 2A = 5\).

Substitute \(B = -1 - A\) from equation 1 into \(B - 2A = 5\):

\(-1 - A - 2A = 5\)

\(-1 - 3A = 5\)

\(-3A = 6\)

\(A = -2\)

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

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

Substitute \(A = -2\) into \(C = -2A\):

\(C = 4\)

Thus, \(\frac{5x - x^2}{(1+x)(2+x^2)} = \frac{-2}{1+x} + \frac{x + 4}{2+x^2}\).

(ii) Expand \(\frac{5x - x^2}{(1+x)(2+x^2)}\) in ascending powers of \(x\):

Expand \(\frac{-2}{1+x}\) using the binomial series:

\(-2(1 - x + x^2 - x^3 + \cdots) = -2 + 2x - 2x^2 + 2x^3 + \cdots\)

Expand \(\frac{x + 4}{2+x^2}\) using the binomial series:

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

Combine the expansions:

\(-2 + 2x - 2x^2 + 2x^3 + \frac{x}{2} + 2 - \frac{x^3}{4} + \cdots\)

Combine like terms:

\(\frac{5}{2}x - 3x^2 + \frac{7}{4}x^3 + \cdots\)

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

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

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

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

Expand and equate coefficients:

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

Equating coefficients gives:

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

Solving these equations, we find \(A = -1\), \(B = 2\), \(C = 1\).

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

(ii) Expand \(f(x)\) in ascending powers of \(x\):

First, expand \(\frac{-1}{1+x}\) using the geometric series:

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

Next, expand \(\frac{2x+1}{1+2x^2}\):

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

Multiply out to get terms up to \(x^3\):

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

Combine the expansions:

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

Simplify to get:

\(f(x) = 3x - 3x^2 - 3x^3\).

(i) Express \(\frac{4 + 5x - x^2}{(1 - 2x)(2 + x)^2}\) in partial fractions:

Assume \(\frac{4 + 5x - x^2}{(1 - 2x)(2 + x)^2} = \frac{A}{1 - 2x} + \frac{B}{2 + x} + \frac{C}{(2 + x)^2}\).

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

\(4 + 5x - x^2 = A(2 + x)^2 + B(1 - 2x)(2 + x) + C(1 - 2x)\).

Expand and equate coefficients to find \(A = 1\), \(B = 1\), \(C = -2\).

(ii) Expand \(\frac{4 + 5x - x^2}{(1 - 2x)(2 + x)^2}\) in ascending powers of \(x\):

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

Expand each term using binomial series:

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

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

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

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

\(1 + \frac{9}{4}x + \frac{15}{4}x^2\).

(a) Express \(f(x)\) in partial fractions:

Assume \(\frac{21 - 8x - 2x^2}{(1 + 2x)(3 - x)^2} = \frac{A}{1 + 2x} + \frac{B}{3 - x} + \frac{C}{(3 - x)^2}\).

Find constants \(A, B, C\) by equating coefficients:

\(A = 2, \ B = 2, \ C = -3\).

(b) Expand each partial fraction:

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

\(\frac{2}{3 - x} = \frac{2}{3}(1 + \frac{x}{3} + \frac{x^2}{9} + \ldots)\)

\(\frac{-3}{(3 - x)^2} = -\frac{1}{3^2}(1 + \frac{2x}{3} + \frac{3x^2}{9} + \ldots)\)

Combine and simplify terms up to \(x^2\):

\(f(x) = \frac{7}{3} - 4x + \frac{215}{27}x^2\).

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

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

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

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

Expand and collect terms:

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

Equate coefficients:

\(A = \frac{2}{3}, \ B = \frac{2}{3}, \ C = \frac{1}{3}\).

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

Use the partial fractions: \(\frac{2/3}{1-x} + \frac{(2/3)x + 1/3}{2+x^2}\).

Expand \(\frac{1}{1-x}\) as \(1 + x + x^2 + \ldots\).

Expand \(\frac{1}{2+x^2}\) as \(\frac{1}{2}(1 - \frac{1}{2}x^2 + \ldots)\).

Multiply and combine terms:

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

Combine and simplify to get:

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

(i) Express \(\frac{5x + 3}{(x + 1)^2(3x + 2)}\) in partial fractions as:

\(\frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{3x+2}\)

Equating and solving for constants, we find \(A = 1, B = 2, C = -3\).

(ii) Expand each term:

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

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

Combine and simplify:

\(\frac{5x + 3}{(x + 1)^2(3x + 2)} \approx \frac{3}{2} - \frac{11}{4}x + \frac{29}{8}x^2\)

(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:

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

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

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

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

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

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

(ii) Show the expansion:

Use the binomial expansion for \((1-x)^{-1}\) and \((1+x^2)^{-1}\):

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

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

Multiply the expansions: \((4x+1)(1-x^2)^{-1}\) and simplify.

Combine terms to obtain: \(f(x) = 3 + 5x - \frac{1}{2}x^2 - \frac{15}{4}x^3\).

(a) Express \(f(x)\) in partial fractions:

Assume \(\frac{2x^2 + 7x + 8}{(1+x)(2+x)^2} = \frac{A}{1+x} + \frac{B}{2+x} + \frac{C}{(2+x)^2}\).

By equating coefficients, we find \(A = 3\), \(B = -1\), \(C = -2\).

(b) Expand each term:

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

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

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

Combine and simplify to get \(2 - \frac{9}{4}x + \frac{5}{2}x^2\).

(a) Express \(f(x)\) in partial fractions:

Assume \(\frac{5x^2 + 8x - 3}{(x-2)(2x^2 + 3)} = \frac{A}{x-2} + \frac{Bx + C}{2x^2 + 3}\).

Multiply through by the denominator: \(5x^2 + 8x - 3 = A(2x^2 + 3) + (Bx + C)(x-2)\).

Equate coefficients to find \(A = 3\), \(B = -1\), \(C = 6\).

(b) Expand \(f(x)\) in ascending powers of \(x\):

Use the expansions:

\((x-2)^{-1} = \frac{1}{2} \left[ 1 - \left( \frac{x}{2} \right) + \left( \frac{x}{2} \right)^2 + \ldots \right]\)

\((2x^2 + 3)^{-1} = \frac{1}{3} \left[ 1 - \frac{2x^2}{3} + \ldots \right]\)

Combine these with the partial fractions:

\(\frac{A}{x-2} = \frac{3}{2} \left[ 1 - \frac{x}{2} + \left( \frac{x}{2} \right)^2 + \ldots \right]\)

\(\frac{Bx + C}{2x^2 + 3} = \frac{-x + 6}{3} \left[ 1 - \frac{2x^2}{3} + \ldots \right]\)

Combine and simplify to get the expansion:

\(\frac{1}{2} - \frac{13}{12}x - \frac{41}{24}x^2\).

(a) Express \(f(x)\) in the form \(\frac{A}{2+x} + \frac{B+Cx}{3+x^2}\).

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

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

(b) Expand \(\left(1 + \frac{1}{2}x\right)^{-1}\) and \(\left(1 + \frac{1}{3}x^2\right)^{-1}\) up to \(x^2\):

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

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

Substitute back into the partial fractions and simplify:

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

Multiply out and collect terms up to \(x^2\):

\(\frac{7}{3} - \frac{5}{3}x + \frac{7}{18}x^2\)

(a) To express \(f(x)\) in partial fractions, assume \(f(x) = \frac{A}{1+2x} + \frac{B}{1-2x} + \frac{C}{2+x}\).

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

\(2 + 11x - 10x^2 = A(1-2x)(2+x) + B(1+2x)(2+x) + C(1+2x)(1-2x)\).

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

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

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

(b) Expand each partial fraction up to \(x^2\):

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

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

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

Add the expansions:

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

Combine like terms:

\(f(x) = 1 + 5x - \frac{7}{2}x^2\).

(a) Express \(f(x)\) in partial fractions:

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

Using a correct method for finding coefficients, we find:

\(A = 1\), \(B = -1\), \(C = 6\).

(b) Obtain the expansion of \(f(x)\):

Use a correct method to find the first two terms of the expansion of each partial fraction:

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

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

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

Substitute back into the partial fractions and simplify:

\(f(x) = (1 + x + x^2) + \left(-\frac{1}{2} - \frac{3}{4}x - \frac{9}{8}x^2\right) + \left(\frac{6}{4} + \frac{18}{4}x + \frac{81}{8}x^2\right)\)

Combine and simplify to get:

\(f(x) = 2 - \frac{11}{4}x + 10x^2\).

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

Assume \(\frac{2x(5-x)}{(3+x)(1-x)^2} = \frac{A}{3+x} + \frac{B}{1-x} + \frac{C}{(1-x)^2}\).

Multiply through by \((3+x)(1-x)^2\) to clear the denominators:

\(2x(5-x) = A(1-x)^2 + B(3+x)(1-x) + C(3+x)\).

Expand and equate coefficients to find \(A = -3\), \(B = -1\), \(C = 2\).

(ii) Expand \(f(x)\) in ascending powers of \(x\):

Use the expansions:

\((3+x)^{-1} = \frac{1}{3} - \frac{1}{9}x + \frac{1}{27}x^2 + \cdots\)

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

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

Substitute into the partial fractions:

\(\frac{-3}{3+x} = -1 + \frac{1}{3}x - \frac{1}{9}x^2 + \cdots\)

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

\(\frac{2}{(1-x)^2} = 2 + 4x + 6x^2 + \cdots\)

Add the expansions and simplify:

\(f(x) = \left(-1 + \frac{1}{3}x - \frac{1}{9}x^2 + \cdots \right) + \left(-1 - x - x^2 + \cdots \right) + \left(2 + 4x + 6x^2 + \cdots \right)\)

Combine like terms:

\(f(x) = \frac{10}{3}x + \frac{44}{9}x^2 + \frac{190}{27}x^3\).