| Date | May 2015 | Marks available | 9 | Reference code | 15M.2.hl.TZ0.3 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Show that | Question number | 3 | Adapted from | N/A |
Question
In 1985 , the deer population in a national park was \(330\). A year later it had increased to \(420\). To model these data the year 1985 was designated as year zero. The increase in deer population from year \(n - 1\) to year \(n\) is three times the increase from year \(n - 2\) to year \(n - 1\). The deer population in year \(n\) is denoted by \({x_n}\).
Show that for \(n \geqslant 2,{\text{ }}{x_n} = 4{x_{n - 1}} - 3{x_{n - 2}}\).
Solve the recurrence relation.
Show using proof by strong induction that the solution is correct.
Markscheme
\({x_n} - {x_{n - 1}} = 3({x_{n - 1}} - {x_{n - 2}})\) M1A2
\({x_n} = 4{x_{n - 1}} - 3{x_{n - 2}}\) AG
we need to solve the quadratic equation \({t^2} - 4t + 3 = 0\) (M1)
\(t = 3,{\text{ }}1\) A1
\({x_n} = a \times {1^n} + b \times {3^n}\)
\({x_n} = a + b \times {3^n}\) A1
\(330 = a + b\) and \(420 = a + 3b\) M1
\(a = 285\) and \(b = 45\) A1
\({x_n} = 285 + 45 \times {3^n}\) A1
\({x_n} = 4{x_{n - 1}} - 3{x_{n - 2}}\)
\({x_n} = 285 + 45 \times {3^n}\)
let \(n = 0 \Rightarrow {x_0} = 330\) A1
let \(n = 1 \Rightarrow {x_1} = 420\) A1
hence true for \(n = 0,{\text{ }}n = 1\)
assume true for \(n = k,{\text{ }}{x_k} = 285 + 45 \times {3^k}\) M1
and assume true for \(n = k - 1,{\text{ }}{x_{k - 1}} = 285 + 45 \times {3^{k - 1}}\) M1
consider \(n = k + 1\)
\({x_{k + 1}} = 4{x_k} - 3{x_{k - 1}}\) M1
\({x_{k + 1}} = 4(285 + 45 \times {3^k}) - 3(285 + 45 \times {3^{k - 1}})\) A1
\({x_{k + 1}} = 4(285) - 3(285) + 4(45 \times {3^k}) - (45 \times {3^k})\) (A1)
\({x_{k + 1}} = 285 + 3(45 \times {3^k})\)
\({x_{k + 1}} = 285 + 45 \times {3^{k + 1}}\) A1
hence if solution is true for \(k\) and \(k - 1\) it is true for. However solution is true for \(k = 0\), \(k = 1\). Hence true for all \(k\). Hence proved by the principle of strong induction R1
Note: Do not award final reasoning mark unless candidate has been awarded at least 4 other marks in this part.
Examiners report
Students often gained full marks on parts a) and b), but a minority of candidates made no start to the question at all.
Students often gained full marks on parts a) and b), but a minority of candidates made no start to the question at all.
In part c) it was pleasing to see a number of fully correct solutions to the strong induction, but many candidates lost marks for not being fully rigorous in the proof.
