Use the recurrence relation to find ${u_2}$.

Find an expression for ${u_n}$ in terms of $n$.

A second recurrence relation, where ${v_1} = {u_1}$ and ${v_2} = {u_2}$, is given by ${v_{n + 1}} + 2{v_n} + {v_{n - 1}} = 0,{\text{ }}n \ge 2$.

Find an expression for ${v_n}$ in terms of $n$.

${u_2} =  - 9$     A1

[1 mark]

METHOD 1

${u_{n + 1}} =  - 2{u_n} - 1$

let ${u_n} = a{( - 2)^n} + b$     M1A1

EITHER

$a{( - 2)^{n + 1}} + b =  - 2\left( {a{{( - 2)}^n} + b} \right) - 1$     M1

$a{( - 2)^{n + 1}} + b = a{( - 2)^{n + 1}} - 2b - 1$

$3b =  - 1$

$b =  - \frac{1}{3}$     A1

${u_1} = 4 \Rightarrow  - 2a - \frac{1}{3} = 4$     (M1)

$\Rightarrow a =  - \frac{{13}}{6}$     A1

OR

using ${u_1} = 4,{\text{ }}{u_2} =  - 9$

$4 =  - 2a + b,{\text{ }} - 9 = 4a + b$     M1A1

solving simultaneously     M1

$\Rightarrow a =  - \frac{{13}}{6},{\text{ and }}b =  - \frac{1}{3}$     A1

THEN

so ${u_n} =  - \frac{{13}}{6}{( - 2)^n} - \frac{1}{3}$

METHOD 2

use of the formula ${u_n} = {a^n}{u_0} + b\left( {\frac{{1 - {a^n}}}{{1 - a}}} \right)$     (M1)

letting ${u_0} =  - \frac{5}{2}$     A1

letting $a =  - 2$ and $b =  - 1$     A1

${u_n} =  - \frac{5}{2}{( - 2)^n} - 1\left( {\frac{{1 - {{( - 2)}^n}}}{{1 -  - 2}}} \right)$     M1A1

$=  - \frac{{13}}{6}{( - 2)^n} - \frac{1}{3}$     A1

[6 marks]

auxiliary equation is ${k^2} + 2k + 1 = 0$     M1

hence $k =  - 1$     (A1)

so let ${v_n} = (an + b){( - 1)^n}$     M1

$(2a + b){( - 1)^2} =  - 9$ and $(a + b){( - 1)^1} = 4$     A1

so $a =  - 5,{\text{ }}b = 1$     M1A1

${v_n} = (1 - 5n){( - 1)^n}$

Note:     Caution necessary to allow FT from (a) to part (c).

[6 marks]

Total [13 marks]