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Which arithmetic sequence explicit formula would yield the following: $1$, $-1$, $1$, $-1$.

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3  
I'd rather say it is geometric, not arithmetic. –  Pedro Tamaroff Nov 20 '12 at 14:43

3 Answers 3

None. A sequence $\,\{a_1,a_2,...\}\,$ is arithmetic iff $\,a_{n+1}-a_n=d=$constant, for any $\,n\geq 1\,$.

In this case it doesn't work, yet your sequence is a geometric one, since

$$\frac{a_{n+1}}{a_n}=-1=\,\text{constant}$$

and thus a general formula for the n-th element is $a_n=1\cdot(-1)^{n-1}=(-1)^{n-1}\,$

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Thanks for your help!!! I appreciate it!! –  Santiago Bueno Nov 20 '12 at 15:12
    
a2, a4, a6,... should be equal to 1, and a1, a3, a5,....should be equal to -1. –  Santiago Bueno Nov 20 '12 at 15:51
    
Indeed so, @Santiago. I wonder why someone thought my answer is incorrect/inappropriate and downvoted me...? Oh, well. –  DonAntonio Nov 20 '12 at 16:57

$a_n = (-1)^{n+1}$ because for odd $n$, you get $1$ and for even $n$, $-1$

for $n \in N$ (I count $N$ as 1, 2,...)

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The sequence is a1 = -1, a2 = 1, a3 = -1, a4 = 1. –  Santiago Bueno Nov 20 '12 at 15:31

the sequence 1, (-1), (-1)(-1), (-1)(-1)(-1), ... is $$(-1)^n$$ for $n = 0,1,2,..$.

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The sequence is a1 = -1, a2 = 1, a3 = -1, a4 = 1. –  Santiago Bueno Nov 20 '12 at 15:23
1  
@SantiagoBueno, so what? also you're mistaken –  sperners lemma Nov 20 '12 at 16:00

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