algebra????/?
hey guys im stuck on these 4 algebra questions i need some help so if you can please help me. 10points and best answer
it says to find the first, fourth, and eighth terms of each sequence.
A(n)=2*3^n-1
A(n)=3*4^n-1
A(n)=3*2^n-1
A(n)=-1*5^n-1
yea i know they are lol that why im asking for help.
*=mulitplication
^= to that power i guess
Whichever term you want to find, you substitute that number for n. (If it’s the first term n = 1, fourth term n = 4 and so on)
I’ll do the first sequence for you and then I’m sure you’ll be able to do the others.
A(n) = 2*3^n-1:
A(1) = 2*3^1-1
A(1) = 2*3^0
A(1) = 2*1 = 2
A(4) = 2*3^4-1
A(4) = 2*3^3
A(4) = 2*9 = 18
A(8) = 2*3^8-1
A(8) = 2*3^7
A(8) = 2*2187 = 4374
September 15th, 2009 at 1:42 am
Whichever term you want to find, you substitute that number for n. (If it’s the first term n = 1, fourth term n = 4 and so on)
I’ll do the first sequence for you and then I’m sure you’ll be able to do the others.
A(n) = 2*3^n-1:
A(1) = 2*3^1-1
A(1) = 2*3^0
A(1) = 2*1 = 2
A(4) = 2*3^4-1
A(4) = 2*3^3
A(4) = 2*9 = 18
A(8) = 2*3^8-1
A(8) = 2*3^7
A(8) = 2*2187 = 4374
References :
I’m a college student who got A’s in Pre-Calc and Calculus
September 15th, 2009 at 2:20 am
I try to do it but i didn’t get the answer and ian algebra 1 student not 2 that’s why I can’t help you sorry
References :
September 15th, 2009 at 2:55 am
A) 5, 161, 13121
B) 11, 767, 196607
C) 5, 47, 767
D) -6, -626, -390626
This took me like fifteen minutes. I hope it helps u. First number is n=1, next is 4, and then 8.
References :