June 2020 Paper 2 Q5
5 Joseph is expanding \((2 - 3x)^7\) in ascending powers of \(x\).
He states that the coefficient of the fourth term is 15120
Joseph’s teacher comments that his answer is almost correct.
Using a suitable calculation, explain the teacher’s comment. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Chooses \(x^3\) term (PI) | B1 | 1.1b |
| Uses correct coefficient formula, allow use of either \({}^7\mathrm{C}_3\) or \({}^7\mathrm{C}_4\) (OE) | M1 | 1.1a |
| Obtains \(-15120\) or 22680 as the value of the coefficient (ignore any power of \(x\) if included as part of the coefficient) | A1F | 1.1b |
| Obtains \(-15120\) and explains how the error could have occurred or what the error is | R1 | 2.4 |
| (4 marks) |
Typical solution
\[(35)(2^4)(-3)^3x^3\]\[-15120x^3\]Joseph has the right number but the wrong sign