Mathematics 9709/13 — May/June 2017

Walkthrough

The question asks for the value of $a$ such that the coefficients of $x$ and $x^2$ in the binomial expansion of $(2 + ax)^7$ are equal. We therefore need to extract the two coefficients from the expansion.

Using the binomial theorem, the term containing $x^r$ is ${}^7C_r (2)^{7-r}(ax)^r$. For the $x$ term, take $r=1$, giving coefficient ${}^7C_1 (2)^6 a$. For the $x^2$ term, take $r=2$, giving coefficient ${}^7C_2 (2)^5 a^2$.

Equating these coefficients gives an equation in $a$. Since the problem states that $a$ is non-zero, we may divide through by $a$ to solve the resulting linear equation. Simplifying the fraction gives $a = \frac{2}{3}$.

Key Takeaways

This question tests the ability to use the binomial theorem to identify individual coefficients without writing out the whole expansion. The key skill is recognizing that the coefficient of $x^r$ is ${}^7C_r (2)^{7-r} a^r$, and that the power of $a$ matches the power of $x$.

Common Mistakes

• Forgetting to include $a^2$ in the coefficient of $x^2$; writing $a$ instead of $a^2$ loses the mark for that coefficient.
• Dividing by $a$ without noting $a \neq 0$, which could introduce the extra solution $a=0$.
• Leaving an extra $x$ in the final answer; the coefficient should be a constant only.
• Not showing the binomial coefficient method, which may lose method marks.

Things to Be Careful About

The mark scheme allows the alternative form $2^7(1 + ax/2)^7$; if using it, equate ${}^7C_1(a/2)$ with ${}^7C_2(a/2)^2$. Ensure the final value is simplified to $\frac{2}{3}$ (or $0.667$). Do not include $a=0$ as an answer because the question specifies a non-zero constant.