Question

A geometric sequence is defined recursively by `a_n = 5a_(n-1)`, the first term of the sequence is 0.45. What is the explicit formula for the nth?

Answer 1

`0.45xx5^(n-1)`

Explanation:

In a geometric series `{a,ar,ar^2,ar^3,.....}`, `a` is first term and ratio is `r`, `n^(th)` term is given by `ar^(n-1)`. Note `r=a_m/a_(m-1)` or `a_m=ra_(m-1)`

As `a_n=5a_(n-1)`, `r=5` and first term `a=0.45`

Hence, `n^(th)` term is given by `0.45xx5^(n-1)`.

Answer 2

`a_n = 5^((n-1)) (0.45)`

Explanation:

The recursive definition `a_n = 5 a_{n-1}` means that

`a_2 = 5 a_1`

`a_color(red)(3) = 5 a_2`

`= 5 * (5 a_1)`

`=5^2 a_1`

`= 5^((color(red)(3)-1)) a_1`

`a_color(red)(4) = 5 a_3`

`= 5 * (5^2 a_1)`

`= 5^3 a_1`

`= 5^((color(red)(4)-1)) a_1`

`...`

We can guess that

`a_color(red)(n) = 5^((color(red)(n)-1)) a_1`

and show that this is true by induction.

For `n = 1`,

`a_1 = 5^((1-1)) a_1 = 5^0 a_1 = a_1`.

For all positive integers `n` greater than 1,

`a_n = 5^((n-1)) a_1`

`= 5 * 5^((n-2)) a_1`

`= 5 * (5^(((n-1)-1)) a_1)`

`= 5 a_{n-1}`.

Since it is given that the first term (`a_1`) is 0.45, the explicit formula for the `n^{"th"}` term is

`a_n = 5^((n-1)) (0.45)`.