Question

A geometric sequence is given by the recursive rule $a_1=\frac{2}{3},\ a_n=3a_{n-1}$ . Give an explicit rule for the nth term of the sequence $b_1,\ b_2,\ b_3,\ \dots\ ,$ given that $b_1=a_1,\ b_2=a_3,\ b_3=a_5,\ \dots$ .

Answer 1

`b_n = (2)/(27) (9^n)`

Explanation:

Given

`a_1=(2)/(3),\ a_n=3a_(n-1).`

Required

An explicit rule for
`b_1,\ b_2,\ b_3,\ \dots`

Where
`b_1=a_1,\ b_2=a_3,\ b_3=a_5,\ \dots`

We have:

`a_1=(2)/(3),\ a_n=3a_(n-1).`

Calculate a2

`a_2 = 3a_(2-1)`

`a_2 = 3a_1`

`a_2 = 3 * (2)/(3)`

`a_2 = 2`

Calculate a3

`a_3 = 3a_(3-1)`

`a_3 = 3a_2`

`a_3 = 3 *2`

`a_3 = 6`

Calculate a4

`a_4 = 3a_(4-1)`

`a_4 = 3a_3`

`a_4 = 3*6`

`a_4 = 18`

Calculate a5

`a_5 = 3a_(5-1)`

`a_5 = 3a_4`

`a_5 = 3*18`

`a_5 = 54`

So:

`b_1=a_1,\ b_2=a_3,\ b_3=a_5,\ \dots`

`a_1=(2)/(3)`
`a_3 = 6`
`a_5 = 54`

The above sequence form a geometric sequence.

Calculate common ratio (r)

`r = (b_3)/(b_2)`

`r = (a_5)/(a_3)`

`r = (54)/(6)`

`r = 9`

So, the explicit formula is:

`b_n = b_1 * r^{n-1`

`b_1=a_1`, so:

`b_n = a_1 * r^{n-1`

`a_1=(2)/(3)`, so:

`b_n = (2)/(3) * 9^{n-1`

Split:

`b_n = (2)/(3) * (9^n)/(9)`

`b_n = (2)/(3*9) (9^n)`

`b_n = (2)/(27) (9^n)`

The explicit rule is:
`b_n = (2)/(27) (9^n)`