Cho dãy số \((u_n)\)xác định bởi: \(u_{1}=\frac{1}{3} \text { và } u_{n+1}=\frac{n+1}{3 n} \cdot u_{n}\). Tổng 1\(S=u_{1}+\frac{u_{2}}{2}+\frac{u_{3}}{3}+. .+\frac{u_{10}}{10}\) bằ...

* Hướng dẫn giải

\(\begin{aligned} &\text { Ta có } u_{n+1}=\frac{n+1}{3 n} u_{n} \Leftrightarrow \frac{u_{n+1}}{n+1}=\frac{1}{3} \cdot \frac{u_{n}}{n}=\frac{1}{3^{n}} u_{1}\\ &\frac{u_{2}}{2}=\frac{1}{3} u_{1} ; \frac{u_{3}}{3}=\frac{1}{3} \frac{u_{2}}{2}=\frac{1}{3^{2}} u_{1} ; \ldots ; \frac{u_{10}}{10}=\frac{1}{3^{9}} u_{1}\\ &\text { Khi đó: }\\ &S=u_{1}+\frac{u_{2}}{2}+\frac{u_{3}}{3}+\ldots+\frac{u_{10}}{10}=\frac{1}{3}+\frac{1}{3^{2}}+\ldots+\frac{1}{3^{10}}=\frac{\frac{1}{3}\left(1-\frac{1}{3^{10}}\right)}{1-\frac{1}{3}}=\frac{29524}{59049} \end{aligned}\)