Cho dãy số $(U_n)$ biết $U_n=\dfrac{2}{n^2+4n+3}$ và dãy $(S_n):\,\left\{ \begin{array}{l} S_1=U_1\\ S_{n+1}=S_n+U_{n+1} \end{array} \right.\,n\in\mathbb{N^*}.$ Xác định công thức tính $(S_n)$ theo $n$.
Thực chất $S_n$ xác định như sau
$S_n = u_{1} + u_{2} + ... + u_{n-1} + u_{n} \quad \forall n \ge 1.$
Ta có
$U_n = \dfrac{2}{(n+1)(n+3)}=\dfrac{1}{n+1}-\dfrac{1}{n+3} \quad \forall n$
do đó
$U_1 =\dfrac{1}{2}-\dfrac{1}{4}$
$U_2 =\dfrac{1}{3}-\dfrac{1}{5}$
 $U_3 =\dfrac{1}{4}-\dfrac{1}{6}$
$\cdots$
$U_{n-1} =\dfrac{1}{n}-\dfrac{1}{n+2}$ 
 $U_{n} =\dfrac{1}{n+1}-\dfrac{1}{n+3}$ 
 Cộng theo từng vế các đẳng thức này ta được
$S_{n} = u_{1} + u_{2} + ... + u_{n-1} + u_{n}=\left (\dfrac{1}{2}+\dfrac{1}{3}+\ldots+\dfrac{1}{n+1} \right )-\left (\dfrac{1}{4}+\dfrac{1}{5}+\ldots+\dfrac{1}{n+3} \right )$
$S_n=\left (\dfrac{1}{2}+\dfrac{1}{3} \right )-\left (\dfrac{1}{n+2}+\dfrac{1}{n+3} \right )=\dfrac{n (13+5 n)}{6 (2+n) (3+n)}\quad \forall n \ge 1.$
Ta có:
$S_n=\sum_{k=1}^n U_n$
      $=\sum_{k=1}^n \frac{2}{n^2+4n+3}$
      $=\sum_{k=1}^n\left(\frac{1}{n+1}-\frac{1}{n+3}\right)$
      $=\frac{5}{6}-\frac{1}{n+2}-\frac{1}{n+3}$

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