Ta có:$S_n=\sum_{k=1}^n U_n$ $=\sum_{k=1}^n \frac{2}{n^2+4n+3}$ $=\sum_{k=1}^n\left(\frac{1}{n+1}-\frac{1}{n+3}\right)$ $=\frac{5}{6}-\frac{1}{n+2}-\frac{1}{n+3}$
Ta có:$S_n=\sum_{k=1}^n U_n$ $=\sum_{k=1}^n \frac{2}{n^2+4n+3}$ $=\sum_{k=1}^n\left(\frac{1}{n+1}-\frac{1}{n+3}\right)$ $=\frac{1}{2}-\frac{1}{n+3}$
Ta có:$S_n=\sum_{k=1}^n U_n$ $=\sum_{k=1}^n \frac{2}{n^2+4n+3}$ $=\sum_{k=1}^n\left(\frac{1}{n+1}-\frac{1}{n+3}\right)$ $=\frac{
5}{6}-\frac{1}{
n+2}-\frac{1}{n+3}$