Hint
Use the whole term (−3x) in each power. The sum of all coefficients is P(1).
Worked solution
(a)
The first terms are \(2^4+4(2^3)(-3x)+6(2^2)(-3x)^2\), giving \(16-96x+216x^2\).
Answer: \(16-96x+216x^2\).
(b)
The coefficient is \(\binom43 2(-3)^3=4(2)(-27)=-216\). The term would be \(\displaystyle -216x^{3}\).
Answer: −216.
(c)
Substituting \(\displaystyle x=1\) adds every coefficient. Thus the sum is \(P(1)=(2-3)^4=1\). The full coefficient list is 16,−96,216,−216,81, which also sums to 1.
Answer: 1.
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 3 marks
- 1 mark: Give the constant term 16.
- 1 mark: Give the linear term −96x.
- 1 mark: Give the quadratic term \(\displaystyle 216x^{2}\).
Alternative methods: Repeated bracket multiplication with the same correctly ordered terms is accepted.
Common error: Omitting the powers of 2 or keeping a negative sign in the squared term.
(b) · 2 marks
- 1 mark: Select the \(\displaystyle r=3\) contribution with its correct constant power and sign.
- 1 mark: Evaluate the coefficient −216.
Alternative methods: Full expansion and correct extraction of the coefficient is accepted.
Common error: Giving a term including \(\displaystyle x^{3}\) when only its coefficient is requested.
(c) · 2 marks
- 1 mark: Use P(1), or add every coefficient of the full correct expansion.
- 1 mark: Give 1.
Alternative methods: Direct coefficient addition from a correct complete expansion is accepted.
Common error: Adding only the first three coefficients from part (a).
Total: 7 marks.