Date | May 2013 | Marks available | 7 | Reference code | 13M.2.hl.TZ2.8 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | Prove | Question number | 8 | Adapted from | N/A |
Question
Use the method of mathematical induction to prove that 52n−24n−1 is divisible by 576 for n∈Z+.
Markscheme
P(n):f(n)=52n−24n−1 is divisible by 576 for n∈Z+
for n=1, f(1)=52−24−1=0
Zero is divisible by 576, (as every non-zero number divides zero), and so P(1) is true. R1
Note: Award R0 for P(1) = 0 shown and zero is divisible by 576 not specified.
Note: Ignore P(2) = 576 if P(1) = 0 is shown and zero is divisible by 576 is specified.
Assume P(k) is true for some k (⇒f(k)=N×576). M1
Note: Do not award M1 for statements such as “let n = k”.
consider P(k+1):f(k+1)=52(k+1)−24(k+1)−1 M1
=25×52k−24k−25 A1
EITHER
=25×(24k+1+N×576)−24k−25 A1
=576k+25×576N which is a multiple of 576 A1
OR
=25×52k−600k−25+600k−24k A1
=25(52k−24k−1)+576k (or equivalent) which is a multiple of 576 A1
THEN
P(1) is true and P(k) true ⇒P(k+1) true, so P(n) is true for all n∈Z+ R1
Note: Award R1 only if at least four prior marks have been awarded.
[7 marks]
Examiners report
This proof by mathematical induction challenged most candidates. While most candidates were able to show that P(1) = 0, a significant number did not state that zero is divisible by 576. A few candidates started their proof by looking at P(2). It was pleasing to see that the inductive step was reasonably well done by most candidates. However many candidates committed simple algebraic errors. The most common error was to state that 52(k+1)=5(5)2k. The concluding statement often omitted the required implication statement and also often omitted that P(1) was found to be true.