Chủ đề này có 1 trả lời

[melodias2002]

Cho dãy số $(x_n)$ xác định bởi: $x_0=2018, x_n=\frac{-2018}{n} \sum_{i=0}^{n-1} x_i (\vee n \geq 1)$. Tính $\lim_{n\rightarrow+\infty} \frac{n^2.\sum_{i=0}^{2018}2^i.x_i+5}{-2019n^2+4n-3}$

[Kim Vu]

Cho dãy số $(x_n)$ xác định bởi: $x_0=2018, x_n=\frac{-2018}{n} \sum_{i=0}^{n-1} x_i (\vee n \geq 1)$. Tính $\lim_{n\rightarrow+\infty} \frac{n^2.\sum_{i=0}^{2018}2^i.x_i+5}{-2019n^2+4n-3}$

Xét $n\leq 2018$
$x_n=\frac{-2018}{n} \sum_{i=0}^{n-1} x_i$

$\Rightarrow  x_{n-1}=\frac{-2018}{n-1} \sum_{i=0}^{n-2} x_i$

$\Rightarrow nx_{n}-(n-1)x_{n-1}=-2018x_{n-1}$

$\Rightarrow x_{n}=\frac{n-2019}{n}x_{n-1}$

$\Rightarrow x_{n}=\frac{n-2019}{n}.\frac{n-2020}{n-1}x_{n-2}$

$=...=x_{0}\frac{(n-2019)(n-2020)...-(2018)}{n(n-1)...1}$

$=x_{0}.\frac{(-1)^n.2018!}{n!.(2018-n)!}$

$=x_{0}.(-1)^nC_{2018}^{n}$

Do đó $\sum_{i=0}^{2018}2^ix_{i}=-2018\sum_{i=0}^{2018}(-1)^iC_{2018}^{i}.2^i=-2018(1-2)^{2018}=-2018$

Vậy $\lim_{n\rightarrow+\infty} \frac{n^2.\sum_{i=0}^{2018}2^i.x_i+5}{-2019n^2+4n-3}=\lim_{n\rightarrow+\infty} \frac{n^2.(-2018)+5}{-2019n^2+4n-3}=\frac{2018}{2019}$