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

Algebra ->  Sequences-and-series -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
.

If you want to have a formula, here it is 


    a%5Bn%5D = sin%28n%2A%28pi%2F2%29%29,  n = 1, 2, 3, 4, 5, 6, 7, 8, . . .