Question

. Find the first five terms of the sequence given by a1 = 2, an = 3an-1 - 1

Answer 1

Given the sequence:

`a_1 =2`

`a_n = 3a_(n-1)-1`

where, n is the number of terms.

We have to find the first five terms of the sequence.

For n = 2;

`a_2 = 3a_(2-1)-1`

⇒
`a_2 = 3a_(1)-1`

Substitute the given values we have;

`a_2 = 3 \cdot 2 -1 = 6-1 = 5`

For n = 3

`a_3 = 3a_(3-1)-1`

⇒
`a_3= 3a_(2)-1`

Substitute the given values we have;

`a_3 = 3 \cdot 5 -1 = 15-1 = 14`

For n = 4

`a_4 = 3a_(4-1)-1`

⇒
`a_4= 3a_(3)-1`

Substitute the given values we have;

`a_4 = 3 \cdot 14 -1 = 42-1 = 41`

For n = 5

`a_5 = 3a_(5-1)-1`

⇒
`a_5= 3a_(4)-1`

Substitute the given values we have;

`a_5= 3 \cdot 41 -1 = 123-1 = 122`

Therefore, the first five terms of the sequence are:

`a_1=2`,
`a_2=5`,
`a_3 =14`,
`a_4 =41` and
`a_5=122`,