Date | May 2017 | Marks available | 6 | Reference code | 17M.1.AHL.TZ1.H_8 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 1 |
Command term | Question number | H_8 | Adapted from | N/A |
Question
Use the method of mathematical induction to prove that 4n+15n−1 is divisible by 9 for n∈Z+.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
let P(n) be the proposition that 4n+15n−1 is divisible by 9
showing true for n=1 A1
iefor n=1, 41+15×1−1=18
which is divisible by 9, therefore P(1) is true
assume P(k) is true so 4k+15k−1=9A, (A∈Z+) M1
Note: Only award M1 if “truth assumed” or equivalent.
consider 4k+1+15(k+1)−1
=4×4k+15k+14
=4(9A−15k+1)+15k+14 M1
=4×9A−45k+18 A1
=9(4A−5k+2) which is divisible by 9 R1
Note: Award R1 for either the expression or the statement above.
since P(1) is true and P(k) true implies P(k+1) is true, therefore (by the principle of mathematical induction) P(n) is true for n∈Z+ R1
Note: Only award the final R1 if the 2 M1s have been awarded.
[6 marks]