a_1&=10
a_(n+1)&=4a_n+1
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)=4a_n+1
4a_n+1
a_(n+1)
-
a_1= 10
-
-
1
a_(1+1)=4a_1+1 ⇕ a_2=4a_1+1
a_2=4 a_1+1 ⇓ a_2=4( 10)+1
a_2= 41
2
a_(2+1)=4a_2+1 ⇕ a_3=4a_2+1
a_3=4 a_2+1 ⇓ a_3=4( 41)+1
a_3= 165
3
a_(3+1)=4a_3+1 ⇕ a_4=4a_3+1
a_4=4 a_3+1 ⇓ a_4=4( 165)+1
a_4= 661
4
a_(4+1)=4a_4+1 ⇕ a_5=4a_4+1
a_5=4 a_4+1 ⇓ a_5=4( 661)+1
a_5= 2645
The first five terms of the sequence are 10, 41, 165, 661, and 2645.
