Question

Cho dãy số xác định bởi u1 = 1; \(u_{n+1}=\dfrac{1}{3}\left(2u_n+\dfrac{n-1}{n^2+3n+2}\right)\). Khi đó \(u_{2030}\) là

Answer 1

\(u_{n+1}=\dfrac{2}{3}u_n+\dfrac{1}{n+2}-\dfrac{2}{3}.\dfrac{1}{n+1}\)

\(\Rightarrow u_{n+1}-\dfrac{1}{n+2}=\dfrac{2}{3}\left(u_n-\dfrac{1}{n+1}\right)\)

Đặt \(u_n-\dfrac{1}{n+1}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=u_1-\dfrac{1}{2}=\dfrac{1}{2}\\v_{n+1}=u_{n+1}-\dfrac{1}{n+2}=\dfrac{2}{3}v_n\end{matrix}\right.\)

\(\Rightarrow v_n\) là cấp số nhân

\(\Rightarrow v_n=\dfrac{1}{2}.\left(\dfrac{2}{3}\right)^{n-1}\)

\(\Rightarrow u_n=v_n+\dfrac{1}{n+1}=\dfrac{1}{2}.\left(\dfrac{2}{3}\right)^{n-1}+\dfrac{1}{n+1}\)