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), the second of which is multiplied by 4.
a_1&=4
a_2&=3x
a_n&=a_(n-1)+4a_(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=a_(n-1)+4a_(n-2)
a_(n-1)+4a_(n-2)
a_n
-
a_1= 4
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-
-
a_2= 3x
-
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3
a_3=a_(3-1)+4a_(3-2) ⇕ a_3=a_2+4a_1
a_3= a_2+4 a_1 ⇓ a_3= 3x+4( 4)
a_3= 3x+16
4
a_4=a_(4-1)+4a_(4-2) ⇕ a_4=a_3+4a_2
a_4= a_3+4 a_2 ⇓ a_4= 3x+16+4( 3x)
a_4= 15x+16
5
a_5=a_(5-1)+4a_(5-2) ⇕ a_5=a_4+4a_3
a_5= a_4+4 a_3 ⇓ a_5= 15x+16+4( 3x+16)
a_5= 27x+80
The first five terms of the sequence are 4, 3x, 3x+16, 15x+16, and 27x+80.
