Cho dãy số (un) xác định bởi $\left\{\begin{array}{l}u_{1}=2 \ u_{n+1}=\frac{4 u_{n}-3}{3 u_{n}-2}\end{array}, \forall n \geq 1\right.$Tính \(A=\li...

Q: Cho dãy số (un) xác định bởi $\left\{\begin{array}{l}u_{1}=2 \\ u_{n+1}=\frac{4 u_{n}-3}{3 u_{n}-2}\end{array}, \forall n \geq 1\right.$Tính \(A=\li...

$u_{n+1}-1=\frac{4 u_{n}-3}{3 u_{n}-2}-1=\frac{u_{n}-1}{3\left(u_{n}-1\right)+1}$Đặt $v_{n}=u_{n}-1 \Rightarrow v_{n+1}=\frac{v_{n}}{3 v_{n}+1} \Rightarrow \frac{1}{v_{n+1}}=\frac{1}{v_{n}}+3$. Vậy $\left(\frac{1}{v_{n}}\right)$ là 1 CSC $d=3, \frac{1}{v_{1}}=1$$\frac{1}{v_{n}}=1+3(n-1)=3 n-2$Nên $\frac{1}{u_{n}-1}=3 n-2$$\begin{array}{l}\text { Ta có } \frac{1}{u_{1}-1}+\frac{1}{u_{1}-2}+\ldots+\frac{1}{u_{n}-1}=3.1-2+3.2-2+\ldots+3 n-2 \\ =3 .(1+2+3+\ldots+n)-2 n\end{array}$$=3 \cdot \frac{n(1+n)}{2}-2 n=\frac{3 n^{2}-n}{2}$Do đó $A=\lim \frac{\frac{1}{u_{1}-1}+\frac{1}{u_{1}-2}+\ldots+\frac{1}{u_{n}-1}}{n^{2}}=\lim \frac{3 n^{2}-n}{2 n^{2}}=\frac{3}{2}$

Question

Accepted Answer

$u_{n+1}-1=\frac{4 u_{n}-3}{3 u_{n}-2}-1=\frac{u_{n}-1}{3\left(u_{n}-1\right)+1}$

Đặt $v_{n}=u_{n}-1 \Rightarrow v_{n+1}=\frac{v_{n}}{3 v_{n}+1} \Rightarrow \frac{1}{v_{n+1}}=\frac{1}{v_{n}}+3$. Vậy $\left(\frac{1}{v_{n}}\right)$ là 1 CSC $d=3, \frac{1}{v_{1}}=1$

$\frac{1}{v_{n}}=1+3(n-1)=3 n-2$

Nên $\frac{1}{u_{n}-1}=3 n-2$

$\begin{array}{l}\text { Ta có } \frac{1}{u_{1}-1}+\frac{1}{u_{1}-2}+\ldots+\frac{1}{u_{n}-1}=3.1-2+3.2-2+\ldots+3 n-2 \ =3 .(1+2+3+\ldots+n)-2 n\end{array}$

$=3 \cdot \frac{n(1+n)}{2}-2 n=\frac{3 n^{2}-n}{2}$

Do đó $A=\lim \frac{\frac{1}{u_{1}-1}+\frac{1}{u_{1}-2}+\ldots+\frac{1}{u_{n}-1}}{n^{2}}=\lim \frac{3 n^{2}-n}{2 n^{2}}=\frac{3}{2}$

Đề thi HSG (CT) 19-20 - Gia Lai - MĐ 6299