Question

For the sequence an = 3a2n – 1 and a0 = 1, what are the values of the first six terms?

Answer 1

The first six terms of the sequence an = 3a2n - 1 with a0 = 1 are 2, 5, 11, 17, 23, and 29.

Step-by-step explanation:

To find the values of the first six terms of the sequence an = 3a2n – 1 with a0 = 1, we can use the given formula to calculate each term.

For n = 0, a0 = 3a(2*0) - 1 = 3a0 - 1 = 3(1) - 1 = 2

For n = 1, a1 = 3a(2*1) - 1 = 3a2 - 1 = 3(2) - 1 = 5

For n = 2, a2 = 3a(2*2) - 1 = 3a4 - 1 = 3(4) - 1 = 11

For n = 3, a3 = 3a(2*3) - 1 = 3a6 - 1 = 3(6) - 1 = 17

For n = 4, a4 = 3a(2*4) - 1 = 3a8 - 1 = 3(8) - 1 = 23

For n = 5, a5 = 3a(2*5) - 1 = 3a10 - 1 = 3(10) - 1 = 29