Mathematics 9709/33 — May/June 2021

Walkthrough

The exponent $\frac{2}{3}$ is not a positive integer, so the finite binomial theorem of Pascal's triangle cannot be used. Instead we use the binomial series for rational $n$:

$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$

In this question the term inside the bracket is $3x$, not $x$, so every occurrence of $x$ on the right-hand side becomes $3x$.

Set $n = \frac{2}{3}$. The first two terms are immediate:

$$1 + nx = 1 + \frac{2}{3}(3x) = 1 + 2x$$

This matches the first mark of the mark scheme.

Next, for the $x^2$ term, apply the coefficient formula:

$$\frac{n(n-1)}{2!}(3x)^2 = \frac{\frac{2}{3}\left(-\frac{1}{3}\right)}{2}(9x^2) = -x^2$$

Here the unsimplified fraction is crucial: simply writing $\binom{2/3}{2}$ is not sufficient for the method mark; the numeric coefficient must be developed.

For the $x^3$ term:

$$\frac{n(n-1)(n-2)}{3!}(3x)^3 = \frac{\frac{2}{3}\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{6}(27x^3) = \frac{4}{3}x^3$$

Collecting the four terms gives the answer.

Key Takeaways

• The binomial theorem applies to fractional $n$, but gives an infinite series.
• Replacing $x$ by $3x$ means each term must include the correct power of $3$.
• The coefficient formula must be applied and simplified with care, especially with negative fractions.

Common Mistakes

• Leaving the coefficient as a symbolic binomial coefficient, such as $\binom{2/3}{2}$, rather than simplifying it. The mark scheme explicitly says this is not enough for the method mark.
• Forgetting to multiply by the powers of $3$ when substituting $3x$.
• Sign errors: $n-1 = -\frac{1}{3}$, so the $x^2$ term is negative.
• Stopping too early: the term in $x^3$ must also be found and simplified.

Things to Be Careful About

• The expansion is valid only when $|3x| < 1$, i.e. $|x| < \frac{1}{3}$, because the exponent is rational.
• Simplify fractions completely: the coefficients should be $2$, $-1$, and $\frac{4}{3}$.
• Remember that the series continues beyond $x^3$; the question only asks for the terms up to and including $x^3$.