a_1&=6
a_(n+1)&=3a_n-n
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-n
3a_n-n
a_(n+1)
-
a_1= 6
-
-
1
a_(1+1)=3a_1- 1 ⇕ a_2=3a_1- 1
a_2=3 a_1-1 ⇓ a_2=3( 6)-1
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= 49
3
a_(3+1)=3a_3- 3 ⇕ a_4=3a_3- 3
a_4=3 a_3-3 ⇓ a_4=3( 49)-3
a_4= 144
4
a_(4+1)=3a_4- 4 ⇕ a_5=3a_4- 4
a_5=3 a_4-4 ⇓ a_5=3( 144)-4
a_5= 428
The first five terms of the sequence are 6, 17, 49, 144, and 428.
