Date | May 2017 | Marks available | 9 | Reference code | 17M.1.AHL.TZ2.H_8 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 2 |
Command term | Prove | Question number | H_8 | Adapted from | N/A |
Question
Prove by mathematical induction that (22)+(32)+(42)+…+(n−12)=(n3), where n∈Z,n⩾3.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(22)+(32)+(42)+…+(n−12)=(n3)
show true for n=3 (M1)
LHS=(22)=1 RHS=(33)=1 A1
hence true for n=3
assume true for n=k:(22)+(32)+(42)+…+(k−12)=(k3) M1
consider for n=k+1:(22)+(32)+(42)+…+(k−12)+(k2) (M1)
=(k3)+(k2) A1
=k!(k−3)!3!+k!(k−2)!2!(=k!3![1(k−3)!+3(k−2)!]) or any correct expression with a visible common factor (A1)
=k!3![k−2+3(k−2)!] or any correct expression with a common denominator (A1)
=k!3![k+1(k−2)!]
Note: At least one of the above three lines or equivalent must be seen.
=(k+1)!3!(k−2)! or equivalent A1
=(k+13)
Result is true for k=3. If result is true for k it is true for k+1. Hence result is true for all k⩾3. Hence proved by induction. R1
Note: In order to award the R1 at least [5 marks] must have been awarded.
[9 marks]