a_1&=-9
a_(n+1)&=2a_n+8
To do so, we will use a table. To find a_2 we will substitute 1 for n in the above formula. To find a_3 we will substitute 2 for n, and so on.
n
a_(n+1)=2a_n+8
2a_n+8
a_(n+1)
-
a_1= -9
-
-
1
a_(1+1)=2a_1+8 ⇕ a_2=2a_1+8
a_2=2 a_1+8 ⇓ a_2=2( -9)+8
a_2= -10
2
a_(2+1)=2a_2+8 ⇕ a_3=2a_2+8
a_3=2 a_2+8 ⇓ a_3=2( -10)+8
a_3= -12
3
a_(3+1)=2a_3+8 ⇕ a_4=2a_3+8
a_4=2 a_3+8 ⇓ a_4=2( -12)+8
a_4= -16
4
a_(4+1)=2a_4+8 ⇕ a_5=2a_4+8
a_5=2 a_4+8 ⇓ a_5=2( -16)+8
a_5= -24
The first five terms of the sequence are -9, -10, -12, -16, and -24.
