## anonymous one year ago 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!

• This Question is Open
1. anonymous

2. freckles

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

3. freckles

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

4. anonymous

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