Question

A geometric sequence is defined by the explicit formula a_n=5(-4)^n-1. what is the recursive formula for the nth term of this sequence?
A- an=-4a_n-1
B-a_n+1=4a_n
C- a_n+1=5a_n
D- a_n=-5a_n-1
whenever there is an underscore(_) it means that the letter should be down, if that makes sense lol.
Will medal and fan!

Answer 1

please help!!!!

Answer 2

\[a_n=5(4)^{n-1} \\ a_{n+1}=5(4)^{n} \\ \text{ also notice } a_n=5(4)^n(4)^{-1} \\ \text{ do you notice what you can replace } 5(4)^n \text{ with ?}\]

Answer 3

and I notice that one number is -4
so replace my 4's with -4's

Answer 4

\[a _{n}=5\left( -4 \right)^{n-1}\]
\[a _{n-1}=5\left( -4 \right)^{n-1-1}=5\left( -4 \right)^{n-1}\left( -4 \right)^{-1}=\frac{ 1 }{ -4 } a {n}\]
\[a _{n}=-4a _{n-1}\]