(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 \(n \in \mathbb{N}\).

(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]

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.