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 6.
a_1&=2
a_2&=x+3
a_n&=a_(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 above formula. To find a_4, we will substitute 4 for n, and so on.
n
a_n=a_(n-1)+6a_(n-2)
a_(n-1)+6a_(n-2)
a_n
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a_1= 2
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-
-
a_2= x+3
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3
a_3=a_(3-1)+6a_(3-2) ⇕ a_3=a_2+6a_1
a_3= a_2+6 a_1 ⇓ a_3= x+3+6( 2)
a_3= x+15
4
a_4=a_(4-1)+6a_(4-2) ⇕ a_4=a_3+6a_2
a_4= a_3+6 a_2 ⇓ a_4= x+15+6( x+3)
a_4= 7x+33
5
a_5=a_(5-1)+6a_(5-2) ⇕ a_5=a_4+6a_3
a_5= a_4+6 a_3 ⇓ a_5= 7x+33+6( x+15)
a_5= 13x+123
The first five terms of the sequence are 2, x+3, x+15, 7x+33, and 13x+123.
