Giair toán nha

Question

Cho

$x(n)$ :

$\begin{cases}x_{1}=1 \\x_{n+1} =\sqrt{x_{n}(x_{n}+1)(x_{n}+2)(x_{n}+3)+1} \end{cases}$

Đặt

$y=\sum_{i=1}^{n}\frac{1}{x_{i}+2}$ . Tính

$\mathop {\lim }\limits yn$

tran85295 · Answer 1 · 3/29/2017 2:12:49 AM

Dễ dàng tìm dc

$x_{n+1}=x_n^2+3x_n+1$

hay

$x_{n+1}+1=(x_n+1)(x_n+2)$

$\Leftrightarrow \frac{1}{x_{n+1}}=\frac{1}{x_n+1}-\frac{1}{x_n+2}\Leftrightarrow \frac{1}{x_n+2}=\frac{1}{x_n+1}-\frac{1}{x_{n+1}+1}$

$\Rightarrow y_n=\sum_{i=1}^n\frac{1}{x_i+2}=\sum_{i=1}^n\left(\frac{1}{x_i+1}-\frac{1}{x_{i+1}+1}\right)=\frac 12-\frac 1{x_{n+1}+1}$

Vì

$x_{n+1}-x_n=(x_n+1)^2 >0 \forall n=1,2,...\Rightarrow x_n$ là dãy tăng

Giả sử

$\lim x_n=\rm L\Rightarrow L=L^2+3L+1\Rightarrow L=-1$ (vô lí)

Vậy

$x_n$ tăng và ko bị chặn

$\Rightarrow \lim x_n=+\infty$

Từ đó ta có

$\lim y_n=\lim\left(\frac 12-\frac{1}{x_{n+1}+1}\right)=\frac 12$

1 Đáp án