Question

[☯๖ۣۜHải☬Ⓢky™]

Tìm tổng S = U₂ + U₃ + ... + U₂₀ ; n ≥ 2

biết U_n = \(\dfrac{2}{\sqrt{n+\sqrt{n^2-4}}}\)

Answer 1

[HT2k02]

\(U_n=\dfrac{2}{\sqrt{n+\sqrt{n^2-4}}}=\dfrac{2\sqrt{2}}{\sqrt{\left(n-2\right)+\left(n+2\right)+2\cdot\sqrt{n+2}\sqrt{n-2}}}\\ =\dfrac{2\sqrt{2}}{\sqrt{\left(\sqrt{n+2}+\sqrt{n-2}\right)^2}}=\dfrac{2\sqrt{2}}{\sqrt{n+2}+\sqrt{n-2}}\\ =\dfrac{\left(n+2\right)-\left(n-2\right)}{\sqrt{2}\left(\sqrt{n+2}+\sqrt{n-2}\right)}=\dfrac{\sqrt{n+2}-\sqrt{n-2}}{\sqrt{2}}\)

\(\Rightarrow S=\dfrac{\sqrt{4}-\sqrt{0}+\sqrt{5}-\sqrt{1}+...+\sqrt{22}-\sqrt{18}}{\sqrt{2}}\\ =\dfrac{\sqrt{19}+\sqrt{20}+\sqrt{21}+\sqrt{22}-\sqrt{0}-\sqrt{1}-\sqrt{2}-\sqrt{3}}{\sqrt{2}}\)