The sequence \(\{ {u_n}\} \) , \(n \in {\mathbb{Z}^ + }\) , satisfies the recurrence relation \({u_{n + 2}} = 5{u_{n + 1}} - 6{u_n}\) .

Given that \({u_1} = {u_2} = 3\) , obtain an expression for \({u_n}\) in terms of n .

The sequence \(\{ {v_n}\} \) , \(n \in {\mathbb{Z}^ + }\) , satisfies the recurrence relation \({v_{n + 2}} = 4{v_{n + 1}} - 4{v_n}\) .

Given that \({v_1} = 2\) and \({v_2} = 12\) , use the principle of strong mathematical induction to show that \({v_n} = {2^n}(2n - 1)\) .

the auxiliary equation is

\({m^2} - 5m + 6 = 0\)     M1

giving \(m = 2,{\text{ 3}}\)     A1

the general solution is

\({u_n} = A \times {2^n} + B \times {3^n}\)     A1

substituting n = 1, 2

\(2A + 3B = 3\)     M1

\(4A + 9B = 3\)     A1

the solution is A = 3, B = –1 giving \({u_n} = 3 \times {2^n} - {3^n}\)     A1

[6 marks]

we first prove that \({v_n} = {2^n}(2n - 1)\) for n = 1, 2     M1

for n = 1, it gives \(2 \times 1 = 2\) which is correct

for n = 2 , it gives \(4 \times 3 = 12\) which is correct     A1

we now assume that the result is true for \(n \leqslant k\)     M1

consider

\({v_{k + 1}} = 4{v_k} - 4{v_{k - 1}}{\text{ }}(k \geqslant 2)\)     M1

\( = {4.2^k}(2k - 1) - {4.2^{k - 1}}(2k - 3)\)     A1

\( = {2^{k + 1}}(4k - 2 - 2k + 3)\)     A1

\( = {2^{k + 1}}\left( {2(k + 1) - 1} \right)\)     A1

this proves that if the result is true for \(n \leqslant k\) then it is true for \(n \leqslant k + 1\)

since we have also proved it true for \(n \leqslant 2\) , the general result is proved by induction     R1

Note: A reasonable attempt has to be made to the induction step for the final R1 to be awarded.

[8 marks]