Verify that the recurrence relation is satisfied by

$${u_n} = A{\alpha ^n} + B{\beta ^n},$$

where $A$ and $B$ are arbitrary constants.

Solve the recurrence relation

${u_{n + 2}} - 2{u_{n + 1}} + 5{u_n} = 0$ given that ${u_0} = 0$ and ${u_1} = 4$.

attempt to substitute the given expression for ${u_n}$ into the recurrence relation     M1

$a{u_{n + 2}} + b{u_{n + 1}} + c{u_n} = a(A{\alpha ^{n + 2}} + B{\beta ^{n + 2}}) + b(A{\alpha ^{n + 1}} + B{\beta ^{n + 1}}) + c(A{\alpha ^n} + B{\beta ^n})$     A1

$= A{\alpha ^n}(a{\alpha ^2} + b\alpha + c) + B{\beta ^n}(a{\beta ^2} + b\beta + c)$     A1

$= 0$ because $\alpha$ and $\beta$ both satisfy $a{x^2} + bx + c = 0$     R1AG

Note:     Award M1A0A1R0 for solutions that are set to zero throughout and conclude with $0 = 0$. Award the R1 for any valid reason.

[4 marks]

the auxiliary equation is ${x^2} - 2x + 5 = 0$     A1

solving their quadratic equation     (M1)

the roots are $1 \pm 2{\text{i}}$     A1

the general solution is

${u_n} = A{\left( {1 + 2{\text{i}}} \right)^n}{\mkern 1mu}  + B{\left( {1 - 2{\text{i}}} \right)^n}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {{u_n} = {{\left( {\sqrt 5 } \right)}^n}\left( {A\,{\text{cis}}\left( {n\arctan 2} \right) + B\,{\text{cis}}\left( { - n\arctan 2} \right)} \right)} \right)$    (A1)

attempt to substitute both boundary conditions     M1

$A + B = 0;{\text{ }}A(1 + 2{\text{i}}) + B(1 - 2{\text{i}}) = 4$     A1

attempt to solve their equations for $A$ and $B$     M1

$A = - {\text{i}},{\text{ }}B = {\text{i}}$     A1

${u_n} = {\text{i}}{(1 - 2{\text{i}})^n} - {\text{i}}{(1 + 2{\text{i}})^n}\,\,\,\left( {{u_n} = 2{{\left( {\sqrt 5 } \right)}^n}\sin (n\arctan 2)} \right)$     A1

Note:     Accept the trigonometric form for ${u_n}$.

[9 marks]