Date | May 2016 | Marks available | 1 | Reference code | 16M.1.hl.TZ0.6 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Conjecture | Question number | 6 | Adapted from | N/A |
Question
Consider the recurrence relation \({H_{n + 1}} = 2{H_n} + 1,{\text{ }}n \in {\mathbb{Z}^ + }\) where \({H_1} = 1\).
Find \({H_2}\), \({H_3}\) and \({H_4}\).
Conjecture a formula for \({H_n}\) in terms of \(n\), for \(n \in {\mathbb{Z}^ + }\).
Prove by mathematical induction that your formula is valid for all \(n \in {\mathbb{Z}^ + }\).
Markscheme
\({H_2} = 2{H_1} + 1\) (M1)
\( = 3;{\text{ }}{H_3} = 7;{\text{ }}{H_4} = 15\) A1
[2 marks]
\({H_n} = {2^n} - 1\) A1
[1 mark]
let \({\text{P}}(n)\) be the proposition that \({H_n} = {2^n} - 1\) for \(n \in {\mathbb{Z}^ + }\)
from (a) \({H_1} = 1 = {2^1} - 1\) A1
so \({\text{P}}(1)\) is true
assume \({\text{P}}(k)\) is true for some \(k \Rightarrow {H_k} = {2^k} - 1\) M1
then \({H_{k + 1}} = 2{H_k} + 1\) (M1)
\( = 2 \times ({2^k} - 1) + 1\) A1
\( = {2^{k + 1}} - 1\)
\({\text{P}}(1)\) is true and \({\text{P}}(k)\) is true \( \Rightarrow {\text{P}}(k + 1)\) is true, hence \({\text{P}}(n)\) is true for all \(n \in {\mathbb{Z}^ + }\) by the principle of mathematical induction R1
Note: Only award the R1 if all earlier marks have been awarded.
[5 marks]
Examiners report
This was a successful question for many candidates with many wholly correct answers seen. The vast majority of candidates were able to answer parts (a) and (b) and most knew how to start part (c). A few candidates were let down by not realising the need for a degree of formality in the presentation of the inductive proof.
This was a successful question for many candidates with many wholly correct answers seen. The vast majority of candidates were able to answer parts (a) and (b) and most knew how to start part (c). A few candidates were let down by not realising the need for a degree of formality in the presentation of the inductive proof.
This was a successful question for many candidates with many wholly correct answers seen. The vast majority of candidates were able to answer parts (a) and (b) and most knew how to start part (c). A few candidates were let down by not realising the need for a degree of formality in the presentation of the inductive proof.