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]