a_1&=5
a_(n+1)&=3a_n+2
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)=3a_n+2
3a_n+2
a_(n+1)
-
a_1= 5
-
-
1
a_(1+1)=3a_1+2 ⇕ a_2=3a_1+2
a_2=3 a_1+2 ⇓ a_2=3( 5)+2
a_2= 17
2
a_(2+1)=3a_2+2 ⇕ a_3=3a_2+2
a_3=3 a_2+2 ⇓ a_3=3( 17)+2
a_3= 53
3
a_(3+1)=3a_3+2 ⇕ a_4=3a_3+2
a_4=3 a_3+2 ⇓ a_4=3( 53)+2
a_4= 161
4
a_(4+1)=3a_4+2 ⇕ a_5=3a_4+2
a_5=3 a_4+2 ⇓ a_5=3( 161)+2
a_5= 485
The first five terms of the sequence are 5, 17, 53, 161, and 485.
