Question

Fie $(1+ \sqrt{3})^n= a_{n}+ b_{n} \sqrt{3}$, ∀n≥1, a(n), b(n)∈ Q. Sa se calculeze $\lim_{n \to \infty} \frac{a_n}{b_n}$. Mersi

Answer 1

Salut,

[tex](1+\sqrt3)^n=a_n+\sqrt3\cdot b_n\ (1)\Rightarrow\\\\\Rightarrow(1-\sqrt3)^n=a_n-\sqrt3\cdot b_n\ (2),\ vezi\;binomul\;lui\;Newton.\\\\Din\ (1)+(2)\;avem\;c\breve{a}:\ a_n=\dfrac{1}2\cdot\left[\left(1+\sqrt3\right)^n+\left(1-\sqrt3\right)^n\right].\\\\Din\ (1)-(2)\;avem\;c\breve{a}:\ b_n=\dfrac{1}{2\sqrt3}\cdot\left[\left(1+\sqrt3\right)^n-\left(1-\sqrt3\right)^n\right].\\\\\dfrac{a_n}{b_n}=\sqrt3\cdot\dfrac{\left(1+\sqrt3\right)^n+\left(1-\sqrt3\right)^n}{\left(1+\sqrt3\right)^n-\left(1-\sqrt3\right)^n}=\sqrt3\cdot\dfrac{1+\left(\dfrac{1-\sqrt3}{1+\sqrt3}\right)^n}{1-\left(\dfrac{1-\sqrt3}{1+\sqrt3}\right)^n}.\\\\\\\dfrac{1-\sqrt3}{1+\sqrt3}=\dfrac{\left(1-\sqrt3\right)^2}{(1-\sqrt3)\cdot(1+\sqrt3)}=\dfrac{4-2\sqrt3}{1-3}=\sqrt3-2.[/tex]

Apoi:

$\dfrac{a_n}{b_n}=\sqrt3\cdot\dfrac{1+\left(\sqrt3-2\right)^n}{1-\left(\sqrt3-2\right)^n}.\\\\Avem\;c\breve{a}:\sqrt3-2\ \textless \ 0,\ apoi\ \sqrt3\ \textgreater \ 1\ |-2\Rightarrow \sqrt3-2\ \textgreater \ -1,\\\\-1\ \textless \ \sqrt3-2\ \textless \ 0,\;deci\;\lim_{\substack{n\to+\infty}}\left(\sqrt3-2\right)^n=0.\\\\La\ final\lim_{\substack{n\to+\infty}}\dfrac{a_n}{b_n}=\sqrt3.$

Green eyes.