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This is a "study guide" question for a test I am trying to figure out. The sequence given is

$a_k=\dfrac{2}{a_{k-1}}$ for all integers $k\ge2$, and $a_1=1$.

It then asks me to find an explicit formula and prove it with "strong" induction. I came up with the following explicit formula:

$a_k=\dfrac{(-1)^k+3}{2}$

But I have no idea how to implement the proof of this. I understand that strong induction is like a normal mathematical induction proof but instead of proving for $k+1$ I need to prove for a range? I'm just not sure what my inductive hypothesis and base cases are for this problem. Thanks for any help!

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    Just use "normal" induction. Strong induction for one proposition is just normal induction for a modified proposition anyway.2012-10-27
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    Well the reason I'm trying to use strong induction is because it says to in the problem - but I have no idea how to apply it to this.2012-10-27
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    Then the problem is stupid. There is no reason to use strong induction here.2012-10-27
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    You *came up* with this formula? The formulas one can come up with when fiddling with the definition, trying some cases and the like, are more likely to resemble $a_{2k}=2$, $a_{2k+1}=1$.2012-10-27
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    @did: What you describe would be considered by many not to constitute _a formula_. Quite possibly OP did find this description and _then_ came up with the mentioned formula.2012-10-27
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    @MarcvanLeeuwen Thanks for your input. I fully disagree that the formula with cases does not constitute a formula, or that it is a less legitimate one than the awkward-looking $\frac12(3+(-1)^n)$. If the OP indeed performed the steps you say, then the step from a formula with cases to a formula without cases but with $(-1)^n$ strikes me as quite useless. Just see how one **uses** these: the first thing one must do is to come back to the description with cases. Ergo.2012-10-27
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    Yep, I think it is an awkward problem all around, but it asked me to find an explicit formula given the sequence, so that it didn't use $k-1$. That's all I did. The point of the problem is to prove it with strong induction, but I was at a loss as to how to apply it here. Seems like that's the general consensus too.2012-10-28

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