Đáp án:
Dãy số đã cho là dãy số giảm
Giải thích các bước giải:
$u_n = \sqrt{n} - \sqrt{3n+1}$
$u_{n+1} = \sqrt{n+1} - \sqrt{3n+4}$
$u_n = \sqrt{n} - \sqrt{3n+1}$
$\Rightarrow u_{n+1} - u_n = \sqrt{n+1} - \sqrt{3n+4} - \sqrt{n} + \sqrt{3n+1}$
$= (\sqrt{n+1} + \sqrt{3n+1})-(\sqrt{3n+4} + \sqrt{n})$
$A = \sqrt{n+1} + \sqrt{3n+1}$
$\Rightarrow A^2 = 4n+2+2\sqrt{3n^2+4n+2}$
$B = \sqrt{3n+4} + \sqrt{n}$
$\Rightarrow B^2=4n+4+2\sqrt{3n^2+4n}$
$C=\sqrt{3n^2+4n+2}$
$\Rightarrow C^2= 3n^2+4n+2$
$=3n^2+4n+1+1$
$D=\sqrt{3n^2+4n} +1$
$\Rightarrow D^2=1+3n^2+4n+2\sqrt{3n^2+4n}$
$= 3n^2+4n+1+2\sqrt{3n^2+4}$
Ta có : $1 <2\sqrt{3n^2+4n}(\forall n \ge 1)$
$\Rightarrow C^2 < D^2$
$\to C < D$
$\Rightarrow \sqrt{3n^2 +4n+2} <1+\sqrt{3n^2+4n}$
$\Rightarrow 2\sqrt{3n^2+4n+2} <2+2\sqrt{3n^2+4n}$
$\Rightarrow 4n+2+2\sqrt{3n^2+4n+2} < 4n+4+2\sqrt{3n^2+4n}$
$\Rightarrow A^2 < B^2$
$\Rightarrow A < B$
$\to u_{n+1} - u_n <0$
