Mathematics 9709/31 — May/June 2011

Walkthrough

We need to expand $\sqrt[3]{(1-6x)}$ as a power series in $x$ up to $x^3$. The standard tool is the binomial expansion, which works for rational indices as well as positive integers:

Here the index is $n = \frac{1}{3}$ and the small quantity is $u = -6x$. Substituting these into the formula gives the four required terms. Each coefficient is then simplified separately: the second term gives $-2x$, the third gives $-4x^2$, and the fourth gives $-\frac{40}{3}x^3$. Adding these together produces the final expansion.

Key Takeaways

• The binomial theorem can be used when the index is a fraction, not just a positive integer.
• Replace $x$ in the standard formula by the whole linear expression, here $-6x$, and simplify carefully.
• The expansion is valid only when the modulus of the substituted term is less than 1, i.e. $|6x| < 1$, or $|x| < \frac{1}{6}$.

Common Mistakes

• Forgetting that the second term is $n(-6x)$, not $n(1-6x)$, and therefore not including the factor $-6$ in every power.
• Sign errors when cubing $-6x$: the cube is negative, so the fourth term must be negative.
• Simplifying the fractional binomial coefficients incorrectly, particularly the fourth coefficient $\frac{5}{81} \cdot (-216) = -\frac{40}{3}$.
• Using only the first two terms and stopping before $x^3$.

Things to Be Careful About

• The mark scheme also accepts a Maclaurin-series method using derivatives; here the binomial route is shown.
• Keep the coefficients simplified: $-4x^2$ not $-\frac{36}{9}x^2$,and $-\frac{40}{3}x^3$ not $-\frac{1080}{81}x^3$.
• The expansion is an approximation valid for small $x$; the question only asks for terms up to $x^3$,so omit higher-order terms.