Question

What is the recursive formula when given the explicit formula for the following geometric sequence?

Answer 1

Answer:

B.

Step-by-step explanation:

When n = 1, we have:

$a_1 = 6\left(-8\right)^{0} = 6\times 1 = 6.$

Now, let's compare $a_{n-1}$ and $a_n$.

We see that: $a_{n-1} = 6\left(-8\right)^{n-2} = 6\left(-8\right)^{n-1}\times \left(-8\right)^{-1}.$

So, finally, substituting $a_n$ and rearranging, we get:

$a_{n-1} = \frac{a_n}{-8} \Rightarrow -8a_{n-1} = a_n.$

In actual fact, this should be expected because the ratio between two terms in the geometric sequence is the number with the exponent!