A sequence $\left\{ {{u_n}} \right\}$ satisfies the recurrence relation ${u_{n + 2}} = 2{u_{n + 1}} - 5{u_n}$ , $n \in \mathbb{N}$ . Obtain an expression for ${u_n}$ in terms of n given that ${u_0} = 0$ and ${u_1} = 1$ .

the auxiliary equation is ${m^2} - 2m + 5 = 0$    M1

the roots are $1 \pm 2{\rm{i}}$    A1

the general solution is ${u_n} = A{\left(1 + 2{\rm{i}}\right)^n} + B{\left(1 - 2{\rm{i}}\right)^n}$     A1

substituting ${u_0} = 0$ and ${u_1} = 1$     M1

$A + B = 0$

$A(1 + 2{\rm{i}}) + B(1 - 2{\rm{i}}) = 1$     A1

Note: Only award above A1 if both equations are correct.

solving

$A\left(1 + 2{\rm{i}} - 1 + 2{\rm{i}}\right) = 1$     (M1)

$A = \frac{1}{{4{\rm{i}}}}$ or $- \frac{{\rm{i}}}{4}$     A1

$B =  - \frac{1}{{4{\rm{i}}}}$ or $\frac{{\rm{i}}}{4}$     A1

therefore

${u_n} = \frac{1}{{4{\rm{i}}}}{\left(1 + 2{\rm{i}}\right)^n} - \frac{1}{{4{\rm{i}}}}{\left(1 - 2{\rm{i}}\right)^n}$ or $\frac{{\rm{i}}}{4}{\left(1 - 2{\rm{i}}\right)^n} - \frac{{\rm{i}}}{4}{\left(1 + 2{\rm{i}}\right)^n}$     A1

[9 marks]