Maclaurin Series: product expansion and binomial coefficient matching — DHS 2025 H2 Math Prelim P1
What this question tests
Question
(a) Find the first three non-zero terms in the Maclaurin series for \(e^x\sin(x+\pi)\).
(b) It is given that the three terms found in part (a) are equal to the first three terms in the series expansion of \(ax(1+bx)^c\) for small \(x\), where \(a\), \(b\) and \(c\) are constants. Find the exact values of \(a\), \(b\) and \(c\). Use these values to find the coefficient of \(x^4\) in the expansion of \(ax(1+bx)^c\), giving your answer as a simplified rational number.
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(a) Maclaurin series for \(e^x\sin(x+\pi)\)
Simplify the sine using \(\sin(x+\pi)=-\sin x\), then multiply the standard series for \(e^x\) and \(\sin x\), keeping terms up to \(x^3\).
\[ \begin{aligned} e^x\sin(x+\pi) &= e^x(-\sin x) = -e^x\sin x \\ &= -\left(1 + x + \frac{x^2}{2!} + \cdots\right)\left(x - \frac{x^3}{3!} + \cdots\right) \\ &= -\left(x - \frac{x^3}{6} + x^2 + \cdots\right) \\ &= -x - x^2 - \frac{x^3}{3} + \cdots \end{aligned} \]\(e^x\sin(x+\pi) = -x - x^2 - \dfrac{x^3}{3} + \cdots\)
(b) Match against \(ax(1+bx)^c\) and find the coefficient of \(x^4\)
Expand \(ax(1+bx)^c\) by the binomial series, compare coefficients with part (a) to solve for \(a\), \(b\), \(c\), then read off the \(x^4\) term.
\[ \begin{aligned} ax(1+bx)^c &= ax\left(1 + cbx + \frac{c(c-1)}{2!}(bx)^2 + \cdots\right) \\ &= ax + abcx^2 + \frac{ab^2c(c-1)}{2}x^3 + \cdots \end{aligned} \]Comparing with \(-x - x^2 - \dfrac{x^3}{3} + \cdots\):
\[ a = -1, \quad abc = -1, \quad \frac{ab^2c(c-1)}{2} = -\frac{1}{3} \]From \(abc = -1\): \(bc = 1 \Rightarrow b = \dfrac{1}{c}\).
From the third equation with \(a=-1\), \(b=\frac{1}{c}\):
\[ \frac{(-1)\times\frac{1}{c^2}\times c(c-1)}{2} = -\frac{1}{3} \implies \frac{c-1}{2c} = \frac{1}{3} \implies 3c-3 = 2c \implies c = 3 \]Therefore \(b = \dfrac{1}{3}\), so \(a = -1\), \(b = \dfrac{1}{3}\), \(c = 3\).
So \(ax(1+bx)^c = -x\left(1+\dfrac{1}{3}x\right)^3\). The \(x^4\) term needs the \(x^3\) term in \(-(1+\frac{1}{3}x)^3\):
\[ -x \times \binom{3}{3}\left(\frac{x}{3}\right)^3 = -x \times \frac{x^3}{27} = -\frac{x^4}{27} \]