Question

5th term for recursive sequence:
a(1) = 1

a(n+1)=3a(n)-1

Answer 1

a(1) = 1

a( n + 1 ) = 3 a ( n ) - 1 = 0

a ( n + 1 ) = 3 a ( n ) - 1

a ( 0 + 1 ) = 3 a ( 0 ) - 1

a ( 1 ) = 3 a ( 0 ) - 1

1 = 3 a ( 0 ) - 1

Add 1 to both sides

1 + 1 = 3 a ( 0 ) - 1 + 1

2 = 3 a ( 0 )

Divide both sides by 3

2 / 3 = a ( 0 )

a ( 0 ) = 2 / 3


n = 1

a ( n + 1 ) = 3 a ( n ) - 1

a ( 1 + 1 ) = 3 a ( 1 ) - 1

a ( 2 ) = 3 ∙ 1 - 1

a ( 2 ) = 3 - 1

a ( 2 ) = 2


n = 2

a ( n + 1 ) = 3 a ( n ) - 1

a ( 2 + 1 ) = 3 a ( 2 ) - 1

a ( 3 ) = 3 ∙ 2 - 1

a ( 3 ) = 6 - 1

a ( 3 ) = 5


n = 3

a ( n + 1 ) = 3 a ( n ) - 1

a ( 3 + 1 ) = 3 a ( 3 ) - 1

a ( 4 ) = 3 ∙ 5 - 1

a ( 4 ) = 15 - 1

a ( 4 ) = 14


n = 4

a ( n + 1 ) = 3 a ( n ) - 1

a ( 4 + 1 ) = 3 a ( 4 ) - 1

a ( 5 ) = 3 ∙ 14 - 1

a ( 5 ) = 42 - 1

a ( 5 ) = 41

Answer 2

a(1) = 1
a(2) = 3a(1)-1 = 3-1 = 2
a(3) = 3a(2)-1 = 6-1 = 5
a(4) = 3a(3)-1 = 15-1 = 14
a(5) = 3a(4)-1 = 42-1 = 41