2015 H2 Mathematics Paper 1 Question 6

Maclaurin Series

Answers

(i)

2x−2x2+83x3{2 x - 2 x^2 + \frac{8}{3} x^3}

(ii)

a=2,  {a=2, \;} b=53,  {b=\frac{5}{3}, \;} c=−35{c=- \frac{3}{5}}
Coefficient of x4=−10427{x^4 = - \frac{104}{27}}

Full solutions

(i)

ln⁡(1+2x)=2x−(2x)22+(2x)33+…=2x−2x2+83x3+…  ■\begin{align*} & \ln ( 1 + 2x ) \\ & = 2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} + \ldots \\ &= 2 x - 2 x^2 + \frac{8}{3} x^3 + \ldots \; \blacksquare \end{align*}

(ii)

ax(1+bx)c=ax(1+cbxax(1++c(c−1)2(bx)2ax(1+++c(c−1)(c−2)3!(bx)3ax(1++++…)=ax+abcx2+ab2c(c−1)2x3+ab3c(c−1)(c−2)3!x4+…\begin{align*} & ax(1+bx)^c \\ &= ax \Bigg( 1 + cbx \\ & \phantom{ax \Bigg( 1 + } + \frac{c(c-1)}{2} (bx)^2 \\ & \phantom{ax \Bigg( 1 + +} + \frac{c(c-1)(c-2)}{3!} (bx)^3 \\ & \phantom{ax \Bigg( 1 + + + } + \ldots \Bigg) \\ &= ax + abc x^2 + \frac{ab^2c(c-1)}{2}x^3 + \frac{ab^3c(c-1)(c-2)}{3!}x^4+ \ldots \end{align*}
Given that the first three terms are equal,
a=2  ■a=2 \; \blacksquare
2bc=−22b2c(c−1)2=83\begin{align} && \quad 2bc &= -2 \\ && \quad \frac{2b^2c(c-1)}{2} &= \frac{8}{3} \\ \end{align}
From (1),{(1),}
b=−1c\begin{align} && \quad b &= -\frac{1}{c} \end{align}
Substituting (3){(3)} into (2):{(2):}
c−1c=833c−3=8cc=−35  ■\begin{align*} \frac{c-1}{c} &= \frac{8}{3} \\ 3c-3 &= 8c \\ c &= - \frac{3}{5} \; \blacksquare \end{align*}
b=53  ■b=\frac{5}{3} \; \blacksquare
Coefficient of x4=ab3c(c−1)(c−2)3!=−10427  ■\begin{align*} & \textrm{Coefficient of } x^4 \\ & = \frac{ab^3c(c-1)(c-2)}{3!} \\ & = - \frac{104}{27} \; \blacksquare \end{align*}