1.Cho $S_{n}$ = $(\sqrt[]{2}+1)^{2}$ + $(\sqrt[]{2}-1)^{2}$ với n = 1; 2 ; 3 ; 4;……. Chứng minh rằng: $S_{2011}$.$S_{2012}$ – $S_{4023}$ = 2$\sq

Ta có

$(\sqrt{2} – 1)(\sqrt{2} + 1) = 2-1 = 1$

Vậy

$\sqrt{2} + 1 = \dfrac{1}{\sqrt{2}-1}=(\sqrt{2}-1)^{-1}$

Khi đó ta có

$S_n = (\sqrt{2}-1)^n + (\sqrt{2}-1)^{-n}= a^n +a^{-n}$

với $a = \sqrt{2}-1$.

Ta tính

$S_{2011}.S_{2012} – S_{4023} = (a^{2011} + a^{-2011})(a^{2012} + a^{-2012}) – (a^{4023} + a^{-4023})$

$= a^{4023} + a^{2011-2012} + a^{-2011+2012} + a^{-4023} – a^{4023} – a^{-4023}$

$= a^{-1} + a^1$

$= a + \dfrac{1}{a}$

$= \sqrt{2}-1 + \dfrac{1}{\sqrt{2}-1}$

$= \sqrt{2}-1 + \sqrt{2} + 1 = 2\sqrt{2} = VP$.

Vậy ta có

$S_{2011}.S_{2012} – S_{4023} = 2\sqrt{2}$