Find the formula for the series: 2,1,0,1,2,1,0,1.....
-1, -1, +1,+1,-1, -1, +1
...directions were "find a formula for the nth term of the *sequence" my mistake
hmmm..dunno how to put it in letters
Any help would be appreciated. Even a shove in the right direction haha.
f(3n+1) = 2, f(3n+2)=1, f(3n)=0
wait, my answer is not right.
f(4n+1) = 2, f(4n+2) = 1, f(4n+3)=0, f(4n)=1.
You could say: f(n) = {2 if n mod 4 = 1, 1 if n mod 4 = 2, 0 if n mod 4 = 3, 1 if n mod 4 = 0}
Ugh. I'm just not seeing it.
How about this: \[f(n) = {1 \over2} (i (-i)^n-i i^n+2)\]
Perfect! But, what does i stand for? :0
well, if you don't have any imagination, there is probably a simpler answer.
I see what you did there! http://i0.kym-cdn.com/photos/images/original/000/105/262/i%20see%20what%20you%20did%20there%202.png
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