find the first five terms of the sequence a_1=5 and a_n+1=3a_(n)+1?

Question

step by step

Answer 1

We have the sequence a_n

We are told that a_1 = 5

Substituting a_1 into a_n+1=3a_(n)+1 will generate a_2:

Let n=1
a(1+1) = 3a_1 + 1
a_(1+1) = 3(5) + 1
a_2 =16

Let n=2
a(2+1) = 3a_2 + 1
a_(2+1) = 3(16) + 1
a_3 = 49

and so on: a_4 is 148 and a_5 is 445.

Answer 2

This problem has issues.
if a_1=5
i) and a_(n+1)=3a_(n+1)
ii) then a_(0+1)=a_1
iii) but a_(n+1)=3a_(n+1)
letting n = 0 and putting this into iii) gives:
iv) a_1=3a_1
this is invalid, unless a_1=0 but you stated that it was equal to 5 which would yield

5=3*5
incorrect

unless it should be read as
a) a_(n+1)=3a(_n) +1

The next issue is what is the stepsize, n does not need to be discreet whole numbers. but I do know that n=0.
Substituting n=0 yields

b) a_(0+1)=3a(0) + 1
c) a_1=3a(_0) + 1
d) (a(_1)-1)/3=a_0
and from the given problem 5=a_1, which can just be substituted in equation (d) to determine what a_0 is.

e) 4/3=a_0

This means the distance from a_0 to a_1 is exactly (5-(4/3))=11/3
so a_2=5+11/3=26/3
a_3=(26/3)+11/3=37/3
a_4=(37/3)+11/3=48/3=16

there ya go
a_0=4/3
a_1=5
a_2=26/3
a_3=37/3
a_4=16
a_5=61/3

Answer 3

a_2 = 3(a_1) + 1 = 3*5 + 1 = 16
a_3 = 3(a_2) + 1 = 3*16 + 1 = 49
a_4 = 3(a_3) + 1 = 3*49 + 1 = 148
a_5 = 3(a_4) + 1 = 3*148 + 1 = 445.
5, 16, 49, 148, 445.

Answer 4

a_2 = 3*5 + 1 = 16
a_3 = 3*16 + 1 =49
etc etc

Answer 5

a_(n+1)=3a_(n)+1?

a_2=3a_1+1
=3*5+1

a_2=16

a_3=3a_2+1
=48+1=49

a_3=49

a_4=3a_3+1
=3*49+1= 147+1=148

a_4=148


a_5=3*a_4+1
=3*148+1

a_5= 445