The given formula means that, after the first two terms of the sequence, every term a_n is the difference between two previous terms a_(n-1) and a_(n-2), both of which are multiplied by 4 and 3 respectively.
a_1&=3
a_2&=2x
a_n&=4a_(n-1)-3a_(n-2)
To do so, we will use a table. To find a_3, we will substitute 3 for n in the above formula. To find a_4, we will substitute 4 for n, and so on.
n
a_n=4a_(n-1)-3a_(n-2)
4a_(n-1)-3a_(n-2)
a_n
-
a_1= 3
-
-
-
a_2= 2x
-
-
3
a_3=4a_(3-1)-3a_(3-2) ⇕ a_3=4a_2-3a_1
a_3=4 a_2-3 a_1 ⇓ a_3=4( 2x)-3( 3)
a_3= 8x-9
4
a_4=4a_(4-1)-3a_(4-2) ⇕ a_4=4a_3-3a_2
a_4=4 a_3-3 a_2 ⇓ a_4=4( 8x-9)-3( 2x)
a_4= 26x-36
5
a_5=4a_(5-1)-3a_(5-2) ⇕ a_5=4a_4-3a_3
a_5=4 a_4-3 a_3 ⇓ a_5=4( 26x-36)-3( 8x-9)
a_5= 80x-117
The first five terms of the sequence are 3, 2x, 8x-9, 26x-36, and 80x-117.
