Question 371452
a<sub>n+2</sub>= 2a<sub>n+1</sub> - 3a<sub>n</sub> where a<sub>0</sub> = a<sub>1</sub> = 3
<pre>
Let n = 0
a<sub>0+2</sub> = 2a<sub>0+1</sub> - 3a<sub>0</sub>
   
a<sub>2</sub> = 2a<sub>1</sub> - 3a<sub>1</sub>
   
a<sub>2</sub> = 2*3 - 3*3 = 6 - 9 = -3

Let n = 1

a<sub>1+2</sub> = 2a<sub>1+1</sub> - 3a<sub>1</sub>

a<sub>3</sub> = 2a<sub>2</sub> - 3a<sub>1</sub>

a<sub>3</sub> = 2*-3 - 3*3 = -6 - 9 = -15

Let n = 2

a<sub>2+2</sub> = 2a<sub>2+1</sub> - 3a<sub>2</sub>

a<sub>4</sub> = 2a<sub>3</sub> - 3a<sub>2</sub>

a<sub>4</sub> = 2*-15 - 3*-3 = -30 + 9 = -21

Keep going and you get:

a<sub>5</sub> = 3
a<sub>6</sub> = 69
a<sub>7</sub> = 129
a<sub>8</sub> = 51
a<sub>9</sub> = -285
a<sub>10</sub> = -723

Edwin</pre>