Date | May 2016 | Marks available | 7 | Reference code | 16M.2.hl.TZ0.5 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Find and Show that | Question number | 5 | Adapted from | N/A |
Question
The sequence \(\{ {u_n}:n \in {\mathbb{Z}^ + }\} \) satisfies the recurrence relation \(2{u_{n + 2}} - 3{u_{n + 1}} + {u_n} = 0\), where \({u_1} = 1,{\text{ }}{u_2} = 2\).
The sequence \(\{ {w_n}:n \in \mathbb{N}\} \) satisfies the recurrence relation \({w_{n + 2}} - 2{w_{n + 1}} + 4{w_n} = 0\), where \({w_0} = 0,{\text{ }}{w_1} = 2\).
(i) Find an expression for \({u_n}\) in terms of \(n\).
(ii) Show that the sequence converges, stating the limiting value.
The sequence \(\{ {v_n}:n \in {\mathbb{Z}^ + }\} \) satisfies the recurrence relation \(2{v_{n + 2}} - 3{v_{n + 1}} + {v_n} = 1\), where \({v_1} = 1,{\text{ }}{v_2} = 2\).
Without solving the recurrence relation prove that the sequence diverges.
(i) Find an expression for \({w_n}\) in terms of \(n\).
(ii) Show that \({w_{3n}} = 0\) for all \(n \in \mathbb{N}\).
Markscheme
(i) the auxiliary equation is \(2{r^2} - 3r + 1 = 0\) (M1)
with roots \(r = 1,{\text{ }}\frac{1}{2}\) A1
the general solution of the difference equation is (M1)
\({u_n} = A + B{\left( {\frac{1}{2}} \right)^n}\) A1
imposing the initial conditions M1
\(A + \frac{B}{2} = 1,{\text{ }}A + \frac{B}{4} = 2\) A1
obtain \({u_n} = 3 - 4{\left( {\frac{1}{2}} \right)^n}\) A1
(ii) as \(n \to \infty ,{\text{ }}{\left( {\frac{1}{2}} \right)^n} \to 0\) R1
\({u_n} \to 3\) A1
hence the sequence is convergent AG
[9 marks]
assume \({v_n} \to L\) M1
taking the limit of both sides of the recurrence relation M1
\(2L - 3L + L{\text{ }}( = 0) = 1\) A1
the contradiction shows that the sequence diverges AG
[3 marks]
(i) the auxiliary equation \({r^2} - 2r + 4 = 0\) A1
has roots \(1 \pm {\text{i}}\sqrt 3 \) A1
METHOD 1
these can be re-expressed as \(2\left( {\cos \left( {\frac{\pi }{3}} \right) \pm {\text{i}}\sin \left( {\frac{\pi }{3}} \right)} \right)\) M1
the general solution is
\({w_n} = {2^n}\left( {A\cos \left( {\frac{{n\pi }}{3}} \right) + B\sin \left( {\frac{{n\pi }}{3}} \right)} \right)\) A1
imposing the initial conditions
\(A = 0,{\text{ }}2B\frac{{\sqrt 3 }}{2} = 2\) A1
obtain \({w_n} = \frac{{{2^{n + 1}}}}{{\sqrt 3 }}\sin \left( {\frac{{n\pi }}{3}} \right)\) A1
METHOD 2
the general solution is
\({w_n} = A{\left( {1 + {\text{i}}\sqrt 3 } \right)^n} + B{\left( {1 - {\text{i}}\sqrt 3 } \right)^n}\) A1
imposing the initial conditions
\(A + B = 0,{\text{ }}A + B + {\text{i}}\sqrt 3 (A - B) = 2\) M1A1
obtain \({w_n} = \frac{1}{{{\text{i}}\sqrt 3 }}{\left( {1 + {\text{i}}\sqrt 3 } \right)^n} - \frac{1}{{{\text{i}}\sqrt 3 }}{\left( {1 - {\text{i}}\sqrt 3 } \right)^n}\) A1
(ii) METHOD 1
\({w_{3n}} = \frac{{{2^{3n + 1}}}}{{\sqrt 3 }}\sin (n\pi )\) R1
\( = 0\) AG
METHOD 2
\({w_{3n}} = \frac{1}{{{\text{i}}\sqrt 3 }}{\left( {1 + {\text{i}}\sqrt 3 } \right)^{3n}} - \frac{1}{{{\text{i}}\sqrt 3 }}{\left( {1 - {\text{i}}\sqrt 3 } \right)^{3n}}\)
\( = \frac{1}{{{\text{i}}\sqrt 3 }}{( - 8)^n} - \frac{1}{{{\text{i}}\sqrt 3 }}{( - 8)^n}\) R1
\( = 0\) AG
[7 marks]
Examiners report
A significant number of candidates had clearly not learned the mechanical procedure for solving linear three-term recurrences. Those who were well prepared, coped well with parts (a) and (c). Part (b) was very rarely successfully answered. Some candidates proved that \({v_{n + 1}} > {v_n}\) but erroneously concluded that the sequence diverged.
A significant number of candidates had clearly not learned the mechanical procedure for solving linear three-term recurrences. Those who were well prepared, coped well with parts (a) and (c). Part (b) was very rarely successfully answered. Some candidates proved that \({v_{n + 1}} > {v_n}\) but erroneously concluded that the sequence diverged.
A significant number of candidates had clearly not learned the mechanical procedure for solving linear three-term recurrences. Those who were well prepared, coped well with parts (a) and (c). Part (b) was very rarely successfully answered. Some candidates proved that \({v_{n + 1}} > {v_n}\) but erroneously concluded that the sequence diverged.