Cho dãy số (un) xác định bởi u1 = 2 và Un=2Un+1-1 với mọi n thuộc N* , có tính chất:

Trả lời:

$u_n=2u_{n+1}-1\Rightarrow u_{n+1}=\frac{u_n+1}{2}$

Ta có:

$u_n=\frac{u_{n-1}+1}{2}$

$\Rightarrow u_{n+1}-u_n=\frac{u_n+1}{2}-u_n$

$=\frac{u_n+1}{2}-\frac{u_{n-1}+1}{2}$

$=\frac{1}{2}\left(u_n-u_{n-1}\right)$

Tương tự ta có: $\Rightarrow u_n-u_{n-1}=\frac{1}{2}\left(u_{n-1}-u_{n-2}\right)$

Tiếp tục như vậy ta được:

$u_{n+1}-u_n=\frac{1}{2}\left(u_n-u_{n-1}\right)$

$u_n-u_{n-1}=\frac{1}{2}\left(u_{n-1}-u_{n-2}\right)$
$u_{n-1}-u_{n-2}=\frac{1}{2}\left(u_{n-2}-u_{n-3}\right)$

….

$u_4-u_3=\frac{1}{2}\left(u_3-u_2\right)$

$u_3-u_2=\frac{1}{2}\left(u_2-u_1\right)$

$\Rightarrow \left(u_{n+1}-u_n\right)\left(u_n-u_{n-1}\right)\left(u_{n-1}-u_{n-2}\right)\mathrm{...}\left(u_4-u_3\right)\left(u_3-u_2\right)$

$=\frac{1}{2}\left(u_n-u_{n-1}\right).\frac{1}{2}\left(u_{n-1}-u_{n-2}\right).\frac{1}{2}\left(u_{n-2}-u_{n-3}\right)\mathrm{...}\frac{1}{2}\left(u_3-u_2\right).\frac{1}{2}\left(u_2-u_1\right)$

$\Rightarrow \left(u_{n+1}-u_n\right)={\left(\frac{1}{2}\right)}^{n-1}.\left(u_2-u_1\right)$

$\iff u_{n+1}-u_n=\frac{1}{2^{n-1}}.\left(u_2-u_1\right)$

Ta có:

$u_1=2u_2-1$

$\Rightarrow u_2=\frac{3}{2}$

$\Rightarrow u_{n+1}-u_n=\frac{1}{2^{n-1}}.\left(\frac{3}{2}-2\right)=-\frac{1}{2^n}<0$

$\Rightarrow \left(u_n\right)$là dãy số giảm

$u_{n+1}-u_n=-\frac{1}{2^n}$

$\iff u_{n+1}=u_n-\frac{1}{2^n}$

Mà $u_n=2u_{n+1}-1$

$\Rightarrow u_n=2\left(u_n-\frac{1}{2^n}\right)-1$

$\iff u_n=2u_n-\frac{1}{2^{n-1}}-1$

$\iff u_n=1+\frac{1}{2^{n-1}}<1+1=2$

$\Rightarrow 1<u_n<2$

Do đó (u_n) là dãy số bị chặn.

Đáp án cần chọn là: B