4
$\begingroup$

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 $n=1$.

$\endgroup$
3
$\begingroup$

$$a_0=\frac{5}{2}$$ $$a_1=\left (\frac{5}{2}\right )^2-2=\frac{17}{4}$$ $$a_2=\left (\left (\frac{5}{2}\right )^2-2\right )^2-2=\left (\frac{17}{4}\right )^2-2=\frac{257}{16}$$
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.

$\endgroup$
7
  • $\begingroup$ Induction proof? (+1) $\endgroup$ – Amad27 Feb 25 '15 at 18:01
  • $\begingroup$ @Amad27 Let me add. Wait a few minutes. $\endgroup$ – AvZ Feb 25 '15 at 18:12
  • $\begingroup$ @Amad27 Took some time to type out the $L_AT_EX$. Sorry about that :P $\endgroup$ – AvZ Feb 25 '15 at 18:26
  • $\begingroup$ perfect.. answer $\endgroup$ – Amad27 Feb 26 '15 at 12:58
  • $\begingroup$ @Amad27 Thanks :) $\endgroup$ – AvZ Feb 26 '15 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.