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If:

$$a_0 = \frac{5}{2}, a_k = a_{k-1}^{2} - 2$$ for $k \ge 1$.

How do I get a general formula for $a_k$? With induction proof. I even tried calculating $a_1, a_2 ...$:

$$a_0 = \frac{5}{2}$$

$$a_1 = \frac{17}{4}$$

$$a_2 = \frac{273}{16}$$

$$a_3 = \frac{74017}{256}$$

I will treat the numerator and denominator seperately.

I see that for the denominator.

$$d = 2^{2^k}$$

Now to the numerator:

I cant get it.

I tried, $$2^{2^{k}} + 1$$ but it doesnt work for $k=2$.

But it is shifted $1$, meaning for $n=2$, I got the value of $n=1$.

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1 Answer 1

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$$a_0=\frac{5}{2}$$ $$a_1=\left (\frac{5}{2}\right )^2-2=\frac{17}{4}$$ $$a_2=\left (\left (\frac{5}{2}\right )^2-2\right )^2-2=\left (\frac{17}{4}\right )^2-2=\frac{257}{16}$$
As you can see, you miscalculated the $a_2$ term and because of that, the next term is also messed up as well. $$a_3=\frac{65537}{256}$$
Note that the numerator is a directly next term of corresponding denominator term (plus one) Therefore, (almost as you predicted), $$a_k=\frac{2^{2^{k+1}}+1}{2^{2^k}}$$

Proof

We can see that
$P(1)$ is true as $$\frac{2^{2}+1}{2^1}=\frac{5}{2}$$
Now, let $P(n)$ be true.
We now have to prove that $P(n+1)$ is true.
$P(n)$ is $$a_{n-1}=\frac{2^{2^n}+1}{2^{2^{n-1}}}$$ $P(n+1)$ is $$a_n=\frac{2^{2^{n+1}}+1}{2^{2^n}}$$
We know that $$a_n=a_{n-1}^2-2$$
If $P(n+1)$ is true, the following will also be true
$$\frac{2^{2^{n+1}}+1}{2^{2^n}}=\left (\frac{2^{2^{n}}+1}{2^{2^{n-1}}}\right )^2-2$$ $$\implies \frac{2^{2^{n+1}}+1}{2^{2^n}}=\frac{(2^{2^{n}}+1)^2}{2^{2^{n}}}-2$$ $$\implies 2^{2^{n+1}}+1=(2^{2^{n}}+1)^2-2^{2^{n}+1}$$ $$\implies 2^{2^{n+1}}+1=2^{2^{n+1}}+2^{2^n+1}+1-2^{2^{n}+1}$$ $$\implies 0=0$$ We can see that $P(n+1)$ is true. Then by induction, QED.

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  • $\begingroup$ Induction proof? (+1) $\endgroup$
    – Amad27
    Feb 25, 2015 at 18:01
  • $\begingroup$ @Amad27 Let me add. Wait a few minutes. $\endgroup$
    – AvZ
    Feb 25, 2015 at 18:12
  • $\begingroup$ @Amad27 Took some time to type out the $L_AT_EX$. Sorry about that :P $\endgroup$
    – AvZ
    Feb 25, 2015 at 18:26
  • $\begingroup$ perfect.. answer $\endgroup$
    – Amad27
    Feb 26, 2015 at 12:58
  • $\begingroup$ @Amad27 Thanks :) $\endgroup$
    – AvZ
    Feb 26, 2015 at 12:59

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