Gọi

D

- Hướng dẫn giải

Đáp án D

có $\frac{4}{n}=\frac{1}{n}+3.\frac{1}{n}$

$\frac{7}{n}=\frac{1}{n}+3.\frac{2}{n}$

$\frac{10}{n}=\frac{1}{n}+3.\frac{3}{n}$

......

$\frac{1+3n}{n}=\frac{1}{n}+3.\frac{n}{n}$

$$\begin{gathered} \Rightarrow S=\frac{1}{n}.n+3(\frac{1}{n}+\frac{2}{n}+\mathrm{...}+\frac{n}{n}) \\ =1+\frac{3}{n}.\frac{1+n}{2}n=1+\frac{3(1+n)}{2} \end{gathered}$$

$\Rightarrow S_{20}=\frac{65}{2}=\mathrm{32,5}$