Giải thích các bước giải:
Câu 12:
Ta có :
$u_n=\dfrac{2}{n^2+4n+3}=\dfrac{2}{(n+1)(n+3)}=\dfrac{1}{n+1}-\dfrac{1}{n+3}$
Vì
$\begin{cases}S_1=u_1\\ S_{n+1}=S_n+u_{n+1}\rightarrow S_{n+1}-S_n=u_{n+1}\end{cases}$
$\rightarrow (S_{n}-S-{n-1})+(S_{n-1}-S_{n-2})+..+S_2-S_1=u_{n}+u_{n-1}+..+u_2$
$\rightarrow S_{n}=u_{n}+u_{n-1}+..+u_2+u_1$
$\rightarrow S_{n}=u_1+u_2+..+u_n$
$\rightarrow S_{n}=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+..+\dfrac{1}{n+1}-\dfrac{1}{n+3}$