Chủ đề này có 4 trả lời

[NLBean]

Tìm $U_{n}$ , biết $U_{n+1} = \sqrt{2} + \sqrt{{U_{n}}^{2} + 1}$ và $U_{1} = 2$

[nhatquangsin]

Ta có:

$U_{n+1}=\sqrt{2}+\sqrt{U_{n}^2+1}=\sqrt{2}+\sqrt{3+U_{n-1}^2+2\sqrt{2U_{n-1}^2+2}}$

$=\sqrt{2}+\sqrt{\frac{(\sqrt{2U_{n-1}^2+2}+2)^2}{2}}=\sqrt{2}+\frac{\sqrt{2U_{n-1}^2+2}+2}{\sqrt{2}}$

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

Vậy $U_{n}=(\sqrt{2})^{n-1}+2$

[nguyenqn1998]

Ta có:

$U_{n+1}=\sqrt{2}+\sqrt{U_{n}^2+1}=\sqrt{2}+\sqrt{3+U_{n-1}^2+2\sqrt{2U_{n-1}^2+2}}$

$=\sqrt{2}+\sqrt{\frac{(\sqrt{2U_{n-1}^2+2}+2)^2}{2}}=\sqrt{2}+\frac{\sqrt{2U_{n-1}^2+2}+2}{\sqrt{2}}$

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

Vậy $U_{n}=(\sqrt{2})^{n-1}+2$

$=>U_{n+1}=U_{1}+n.\sqrt{2}=2+n.\sqrt{2}$

[nhatquangsin]

$=>U_{n+1}=U_{1}+n.\sqrt{2}=2+n.\sqrt{2}$

uk mình bị nhầm

[libach80]

Ta có:

$U_{n+1}=\sqrt{2}+\sqrt{U_{n}^2+1}=\sqrt{2}+\sqrt{3+U_{n-1}^2+2\sqrt{2U_{n-1}^2+2}}$

$=\sqrt{2}+\sqrt{\frac{(\sqrt{2U_{n-1}^2+2}+2)^2}{2}}=\sqrt{2}+\frac{\sqrt{2U_{n-1}^2+2}+2}{\sqrt{2}}$

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

Vậy $U_{n}=(\sqrt{2})^{n-1}+2$

Ta có $u_n=\sqrt{2}+\sqrt{u^{2}_{n-1}+1}\rightarrow u_{n}^{2}+1=4+u^{2}_{n-1}+2\sqrt{2u^{2}_{n-1}+2}$ khi đó dòng đầu tiên ko đúng?