The function $f(x) = \frac{{1 + ax}}{{1 + bx}}$ can be expanded as a power series in x, within its radius of convergence R, in the form $f(x) \equiv 1 + \sum\limits_{n = 1}^\infty  {{c_n}{x^n}}$ .

(a)     (i)     Show that ${c_n} = {( - b)^{n - 1}}(a - b)$.

(ii)     State the value of R.

(b)     Determine the values of a and b for which the expansion of f(x) agrees with that of ${{\text{e}}^x}$ up to and including the term in ${x^2}$ .

(c)     Hence find a rational approximation to ${{\text{e}}^{\frac{1}{3}}}$ .

(a)     (i)     $f(x) = (1 + ax){(1 + bx)^{ - 1}}$

$= (1 + ax)(1 - bx + ...{( - 1)^n}{b^n}{x^n} + …$     M1A1

it follows that

${c_n} = {( - 1)^n}{b^n} + {( - 1)^{n - 1}}a{b^{n - 1}}$     M1A1

$= {( - b)^{n - 1}}(a - b)$     AG

(ii)     ${\text{R}} = \frac{1}{{\left| b \right|}}$     A1

[5 marks]

(b)     to agree up to quadratic terms requires

$1 = - b + a,{\text{ }}\frac{1}{2} = {b^2} - ab$     M1A1A1

from which $a = - b = \frac{1}{2}$     A1

[4 marks]

(c)     ${{\text{e}}^x} \approx \frac{{1 + 0.5x}}{{1 - 0.5x}}$     A1

putting $x = \frac{1}{3}$     M1

${{\text{e}}^{\frac{1}{3}}} \approx \frac{{\left( {1 + \frac{1}{6}} \right)}}{{\left( {1 - \frac{1}{6}} \right)}} = \frac{7}{5}$     A1

[3 marks]

Total [12 marks]

Most candidates failed to realize that the first step was to write f(x) as $(1 + ax){(1 + bx)^{ - 1}}$ . Given the displayed answer to part(a), many candidates successfully tackled part(b). Few understood the meaning of the ‘hence’ in part(c).