Compare coefficients without full expansions · source reference

Printed question

The coefficient of $x^3$ in the expansion of $\frac1{(2+ax)^5}$ is 5 times the coefficient of $x^2$ in the expansion of $\sqrt{4-ax}$ where $a$ is a non-zero constant. Find the exact value of $a$. [5]

[2020/HCI/Prelim/I/2]

Printed working

$$(2+ax)^{-5}=2^{-5}(1+ax/2)^{-5}=\frac1{32}[1-5ax/2+\tfrac{(-5)(-6)}{2!}(ax/2)^2+\tfrac{(-5)(-6)(-7)}{3!}(ax/2)^3+\cdots].$$

$[x^3]=\frac1{32}\frac{(-5)(-6)(-7)}{3!}(a/2)^3=-35a^3/256$.

$\sqrt{4-ax}=2(1-ax/4)^{1/2}=2[1+\frac12(-ax/4)+\frac{(1/2)(-1/2)}{2!}(-ax/4)^2+\cdots]$.

$[x^2]=2\frac{(1/2)(-1/2)}{2!}(-a/4)^2=-a^2/64$.

$-35a^3/256=5(-a^2/64)\Rightarrow35a^3=20a^2\Rightarrow7a=4$ since $a\ne0$, hence $a=4/7$.

Mathematical review

$[x^3](2+ax)^{-5}=2^{-5}\binom{-5}{3}(a/2)^3=-35a^3/256$. Also $[x^2]\sqrt{4-ax}=2\binom{1/2}{2}(-a/4)^2=-a^2/64$.

Hence $-35a^3/256=5(-a^2/64)$. Since $a\ne0$, cancellation gives $35a=20$, so $a=4/7$.

Provenance

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