The given formula means that, after the first two terms of the sequence, every term a_n is the sum of two previous terms a_(n-1) and a_(n-2), both of which are multiplied by 3 and 6 respectively.
a_1&=1
a_2&=x
a_n&=3a_(n-1)+6a_(n-2)
To do so, we will use a table. To find a_3, we will substitute 3 for n in the formula above. To find a_4, we will substitute 4 for n, and so on.
n
a_n=3a_(n-1)+6a_(n-2)
3a_(n-1)+6a_(n-2)
a_n
-
a_1= 1
-
-
-
a_2= x
-
-
3
a_3=3a_(3-1)+6a_(3-2) ⇕ a_3=3a_2+6a_1
a_3=3 a_2+6 a_1 ⇓ a_3=3 x+6( 1)
a_3= 3x+6
4
a_4=3a_(4-1)+6a_(4-2) ⇕ a_4=3a_3+6a_2
a_4=3 a_3+6 a_2 ⇓ a_4=3( 3x+6)+6 x
a_4= 15x+18
5
a_5=3a_(5-1)+6a_(5-2) ⇕ a_5=3a_4+6a_3
a_5=3 a_4+6 a_3 ⇓ a_5=3( 15x+18)+6( 3x+6)
a_5= 63x+90
The first five terms of the sequence are 1, x, 3x+6, 15x+18, and 63x+90.
