Recover parameters and a general coefficient · source reference

Printed question

It is given that the first 3 terms in the expansion of $(2-ax)^{-b}$ in ascending powers of $x$ are $\frac18+\frac9{16}x+cx^2+\cdots$, where $a,b,c$ are positive constants.

(i) Find the values of $a,b,c$. [7]

(ii) Show that the coefficient of $x^r$ is in the form $p(r+1)(r+2)q^r$, where $p$ and $q$ are constants to be determined. [4]

[ACJC/2009/Prelim/II/2]

Printed working

$$(2-ax)^{-b}=2^{-b}(1-ax/2)^{-b}=2^{-b}[1+(-b)(-ax/2)+\tfrac{(-b)(-b-1)}{2!}(-ax/2)^2+\cdots]=2^{-b}[1+abx/2+a^2b(b+1)x^2/8+\cdots].$$

Constant: $2^{-b}=1/8\Rightarrow b=3$. Linear: $(1/8)(ab/2)=9/16\Rightarrow a=3$. Quadratic: $c=(1/8)(a^2b(b+1)/8)=27/16$.

$$\begin{aligned}[x^r]&=2^{-b}\frac{(-b)(-b-1)\cdots(-b-(r-1))}{r!}(-a/2)^r\\&=\tfrac18\frac{(-3)(-4)\cdots(-r-2)}{r!}(-3/2)^r\\&=\tfrac18(-1)^{2r}\frac{3\cdot4\cdots(r+2)}{r!}(3/2)^r\\&=\tfrac18\frac{(r+1)(r+2)}2(3/2)^r=\tfrac1{16}(r+1)(r+2)(3/2)^r.\end{aligned}$$

Thus $p=1/16$, $q=3/2$.

Teacher note 1

Need to factorise out $2$ to make the constant term 1. Use the law of indices $(AB)^n=A^nB^n$. Then use the binomial theorem

$$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\cdots,$$

where $n=-b$ and replace $x$ by $-ax/2$.

Teacher note 2

In the coefficient of $x^r$, $(-b-0)(-b-1)\cdots(-b-(r-1))/r!$: there are $r$ factors in the numerator for $r!$ in the denominator.

Mathematical review

$(2-ax)^{-b}=2^{-b}[1+(ba/2)x+b(b+1)a^2x^2/8+\cdots]$. The constant gives $b=3$; $(1/8)(3a/2)=9/16$ gives $a=3$. Then $c=(1/8)(3\cdot4\cdot9/8)=27/16$.

Now $(2-3x)^{-3}=\frac18(1-3x/2)^{-3}$. For integer $r\ge0$, $[x^r]=\frac18\binom{r+2}{2}(3/2)^r=\frac1{16}(r+1)(r+2)(3/2)^r$. Thus $p=1/16,q=3/2$; the series is valid for $|x|<2/3$.

Provenance

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