| Date | May 2017 | Marks available | 4 | Reference code | 17M.2.hl.TZ0.3 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Determine | Question number | 3 | Adapted from | N/A |
Question
The sequence \(\{ {u_n}\} \) satisfies the second-degree recurrence relation
\[{u_{n + 2}} = {u_{n + 1}} + 6{u_n},{\text{ }}n \in {\mathbb{Z}^ + }.\]
Another sequence \(\{ {v_n}\} \) is such that
\[{v_n} = {u_{2n}},{\text{ }}n \in {\mathbb{Z}^ + }.\]
Write down the auxiliary equation.
Given that \({u_1} = 12,{\text{ }}{u_2} = 6\), show that
\[{u_n} = 2 \times {3^n} - 3 \times {( - 2)^n}.\]
Determine the value of \(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n} + {u_{n - 1}}}}{{{u_n} - {u_{n - 1}}}}\).
Determine the second-degree recurrence relation satisfied by \(\{ {v_n}\} \).
Markscheme
the auxiliary equation is \({m^2} - m - 6 = 0\) or equivalent A1
[??? marks]
attempt to solve quadratic (M1)
the roots are \(3,{\text{ }} - 2\) A1
the general solution is
\({u_n} = A \times {3^n} + B \times {( - 2)^n}\) A1
initial conditions give
\(3A - 2B = 12\)
\(9A + 4B = 6\) M1
the solution is \(A = 2,{\text{ }}B = - 3\) A1
\({u_n} = 2 \times {3^n} - 3 \times {( - 2)^n}\) AG
[??? marks]
\({u_n} + {u_{n - 1}} = 2 \times {3^n} - 3 \times {( - 2)^n} + 2 \times {3^{n - 1}} - 3 \times {( - 2)^{n - 1}}\) M1
\( = 8 \times {3^{n - 1}} + {\text{multiple of }}{2^{n - 1}}\) A1
\({u_n} - {u_{n - 1}} = 2 \times {3^n} - 3 \times {( - 2)^n} - 2 \times {3^{n - 1}} + 3 \times {( - 2)^{n - 1}}\)
\( = 4 \times {3^{n - 1}} + {\text{multiple of }}{2^{n - 1}}\) A1
any evidence of noting that the \({3^{n - 1}}\) terms dominate R1
\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n} + {u_{n - 1}}}}{{{u_n} - {u_{n - 1}}}} = 2\) A1
[??? marks]
\({v_n} = 2 \times {3^{2n}} - 3 \times {( - 2)^{2n}}\) M1
\( = 2 \times {9^n} - 3 \times {4^n}\) A1
the auxiliary equation is
\({m^2} - 13m + 36 = 0\) A1
the recurrence relation is
\({v_{n + 2}} = 13{v_{n + 1}} - 36{v_n}\) A1
[4 marks]
