Mathematics 9709/12 — October/November 2017

Walkthrough

A binomial expansion of $(a+b)^9$ has terms of the form ${}^{9}\mathrm{C}_r a^{9-r}b^r$. Here $a = 2x$ and $b = -\frac{1}{4x^2}$. The term independent of $x$ is the one where all powers of $x$ cancel, so we first write the general term and collect the exponent of $x$.

The factor $(2x)^{9-r}$ contributes $x^{9-r}$, while $\left(-\frac{1}{4x^2}\right)^r$ contributes $x^{-2r}$. Together the power of $x$ is $x^{9-3r}$. Setting $9-3r=0$ gives $r=3$, so we need the fourth term of the expansion.

Now evaluate the coefficient. The binomial coefficient is ${}^{9}\mathrm{C}_3 = 84$. The constant from $(2x)^{6}$ is $2^6 = 64$. The constant from $\left(-\frac{1}{4x^2}\right)^3$ is $\left(-\frac14\right)^3 = -\frac{1}{64}$. Multiplying gives $84 \times 64 \times \left(-\frac{1}{64}\right) = -84$.

Key Takeaways

This question tests the general term of a binomial expansion and the idea of matching powers. To find a term independent of $x$, set the total power of $x$ in the general term equal to zero and solve for $r$. It also checks careful handling of negative fractional constants.

Common Mistakes

• Choosing $r = 0$ because the term is described as independent; the correct $r$ is found by setting the $x$-power to zero, not by inspection.
• Forgetting the binomial coefficient ${}^{9}\mathrm{C}_3$.
• Getting the sign wrong: $\left(-\frac14\right)^3$ is negative.
• Using $2^{3}$ instead of $2^{6}$ because $9-r = 6$ when $r=3$.
• Forgetting that the denominator $x^2$ contributes a negative exponent $-2r$.

Things to Be Careful About

• The mark scheme requires the selected term to be shown, not just the final answer; write the full coefficient expression.
• Combine the powers of $x$ before deciding $r$.
• Keep the negative sign with the fraction $\frac{1}{4x^2}$ throughout.
• Simplify $84 \times 64 \times \left(-\frac{1}{64}\right)$ carefully; the $64$ cancels with the denominator $64$.