SOLUTION: Hello...what rule could be here? 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0... i know if would be 1 0 1 0 1 0 1 0 1 0 would be a_n=1/2((-1)^(n+1)+1) and -1 0 -1 0 -1 0 -1 0 -1 0 -1 0

Question 1167820: Hello...what rule could be here?
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0...
i know if would be 1 0 1 0 1 0 1 0 1 0 would be a_n=1/2((-1)^(n+1)+1) and -1 0 -1 0 -1 0 -1 0 -1 0 -1 0 a_n=1/2((-1)^(n+1)-1) but being stuck in finding solution for 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0... could you help me?
Answer by ikleyn(52788) (Show Source): You can put this solution on YOUR website!
.


If you want to have a formula, here it is 


     = ,  n = 1, 2, 3, 4, 5, 6, 7, 8, . . .

RELATED QUESTIONS
Hi,my name is Natalia. I solved two problems, but I'm not sure that I did it right. I... (answered by venugopalramana)

2 sin2 0 -1 =... (answered by greenestamps)

1 2 0 -2 -2 0 0 -1 -1 (answered by Alan3354)

Let A = [0 0 1] [8 1 0] [2 0 0] Find Y such that YA = [2 0 0] [8 1 0] [0 (answered by Fombitz)

i'm unable to understand how to sketch a digraph? of matrices its so confusing 1) 0 1 (answered by Fombitz,violettta222)

Hi all, I was hoping someone would explain to me how to write the matrix below (B) in... (answered by Fombitz)

A =[1 0 -1 0 0 0 ] B= [-2 -1 0 3] C= [1 0... (answered by ikleyn)

0 cos=1 0... (answered by chessace)

determinant of the matrix 1 2 -1 -2 1 2 1 2 1 0 0 1 0 1 1... (answered by khwang)