October 2020 Paper 2 Q4

Notes

M1: For an attempt at the correct coefficient of \(x^4\).
The coefficient must have

• the correct binomial coefficient
• the correct power of \(a\)
• 2 or \(2^4\) (may be implied)

May be seen within a full or partial expansion.
Accept \({}^{7}\mathrm{C}_{4}a^3(2x)^4,\ \dfrac{7!}{4!3!}a^3(2x)^4,\ \dbinom{7}{4}a^3(2x)^4,\ 35a^3(2x)^4,\ 560a^3x^4,\ \dbinom{7}{4}a^3 16x^4\) etc.
or \({}^{7}\mathrm{C}_{4}a^3 2^4,\ \dfrac{7!}{4!3!}a^3 2^4,\ \dbinom{7}{4}a^3 2^4,\ 35a^3 2^4,\ 560a^3\) etc.
or \({}^{7}\mathrm{C}_{3}a^3(2x)^4,\ \dfrac{7!}{4!3!}a^3(2x)^4,\ \dbinom{7}{3}a^3(2x)^4,\ 35a^3(2x)^4,\ 560a^3x^4,\ \dbinom{7}{3}a^3 16x^4\) etc.
or \({}^{7}\mathrm{C}_{3}a^3 2^4,\ \dfrac{7!}{4!3!}a^3 2^4,\ \dbinom{7}{3}a^3 2^4,\ 35a^3 2^4,\ 560a^3\)
You can condone missing brackets around the “\(2x\)” so allow e.g. \(\dfrac{7!}{4!3!}a^3 2x^4\)
An alternative is to attempt to expand \(a^7\left(1 + \dfrac{2x}{a}\right)^7\) to give \(a^7\left(\ldots\dfrac{7 \times 6 \times 5 \times 4}{4!}\left(\dfrac{2x}{a}\right)^4\ldots\right)\)
Allow M1 for e.g. \(a^7\left(\ldots\dfrac{7 \times 6 \times 5 \times 4}{4!}\left(\dfrac{2x}{a}\right)^4\ldots\right),\ a^7\left(\ldots\dbinom{7}{4}\left(\dfrac{2x}{a}\right)^4\ldots\right),\ a^7\left(\ldots 35\left(\dfrac{2x}{a}\right)^4\ldots\right)\) etc.
but condone missing brackets around the \(\dfrac{2x}{a}\)
Note that \({}^{7}\mathrm{C}_{3},\ \dbinom{7}{3}\) etc. are equivalent to \({}^{7}\mathrm{C}_{4},\ \dbinom{7}{4}\) etc. and are equally acceptable.
If the candidate attempts \((a + 2x)(a + 2x)(a + 2x)\ldots\) etc. then it must be a complete method to reach the required term. Send to review if necessary.

dM1: For \(\text{``}560\text{''}a^3 = 15120 \Rightarrow a = \ldots\) Condone slips on copying the 15120 but their “560” must be an attempt at \({}^{7}\mathrm{C}_{4} \times 2\) or \({}^{7}\mathrm{C}_{4} \times 2^4\) and must be attempting the cube root of \(\dfrac{15120}{\text{``}560\text{''}}\). Depends on the first mark.

A1: \(a = 3\) and no other values i.e. \(\pm 3\) scores A0

Note that this is fairly common:
\({}^{7}\mathrm{C}_{4}a^3 2x^4 = 70a^3x^4 \Rightarrow 70a^3 = 15120 \Rightarrow a^3 = 216 \Rightarrow a = 6\)
and scores M1 dM1 A0