Theo đề, ta có: $\frac{1}{u_{n}-1}=-\frac{1}{2}(\frac{2}{5})^n-1/2$
$\Rightarrow S_{10}=\sum_{n=1}^{10}\frac{1}{u_{n}-1}=-\frac{1}{2}\sum_{n=1}^{10}(\frac{2}{5})^{n}-5=\frac{(2/5)((2/5)^{10}-1)}{2(1-2/5)}-5=\frac{1}{3}(2^{10}/5^{10}-1)-5$
- Nam Antoneus yêu thích