<s>[INST] What is the pattern in the powers of $\sqrt{2}-\sqrt{1}$? What is the pattern in this?
$$\begin{align}
\left(\sqrt{2}-\sqrt{1}\right)^1 &= \sqrt{2}-\sqrt{1}\\
\left(\sqrt{2}-\sqrt{1}\right)^2 &= \sqrt{9}-\sqrt{8}\\
\left(\sqrt{2}-\sqrt{1}\right)^3 &= \sqrt{50}-\sqrt{49}\\
\left(\sqrt{2}-\sqrt{1}\right)^4 &= \sqrt{289}-\sqrt{288}\\
\end{align}$$
I thought of applying the binomial theorem
 [/INST] Given that
$$\left(\sqrt{2}-\sqrt{1}\right)^k = \sqrt{A_k}-\sqrt{A_k-1}$$
then
$$\left(\sqrt{2}-\sqrt{1}\right)^{k+1} = \left(\sqrt{2}-\sqrt{1}\right)(\sqrt{A_k}-\sqrt{A_k-1})=\\=\sqrt 2\sqrt{A_k}-\sqrt 2\sqrt{A_k-1}-\sqrt{A_k}+\sqrt{A_k-1}=$$
$$=\sqrt{A_{k+1}}-\sqrt{A_{k+1}-1}$$
with

*

*$\sqrt{A_{k+1}}=\sqrt 2\sqrt{A_k}+\sqrt{A_k-1} \implies A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)}$


*$\sqrt{A_{k+1}-1}=\sqrt 2\sqrt{A_k-1}+\sqrt{A_k}\implies A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)}$
that is

$$A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)}$$

and $A_1=2$.

Using an approach similar to this one, we have
$$A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)} \\\iff 2A_{k+1}-1=3(2A_{k}-1)+2\sqrt2\sqrt{(2A_k-1)^2-1}$$
and by $2A_{k+1}-1=\frac12\left(t_k+\frac1{t_k}\right) \implies t_1=3+2\sqrt 2$ we obtain
$$t_{k+1}+\frac1{t_{k+1}}=3\left(t_k+\frac1{t_k}\right)+2\sqrt 2\sqrt{\left(t_k+\frac1{t_k}\right)^2-4}$$
$$t_{k+1}+\frac1{t_{k+1}}=3\left(t_k+\frac1{t_k}\right)+2\sqrt 2\left(t_k-\frac1{t_k}\right)$$
$$t_{k+1}+\frac1{t_{k+1}}=(3+2\sqrt 2)t_k+\frac{3-2\sqrt 2}{t_k}$$
$$t_{k+1}+\frac1{t_{k+1}}=(3+2\sqrt 2)t_k+\frac{1}{(3+2\sqrt 2)t_k}$$
$$t_{k}=(3+2\sqrt 2)^k$$
and then

$$A_k=\frac{(3+2\sqrt 2)^k+(3-2\sqrt 2)^k+2}{4}$$

</s>