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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
$S_{n} = u_{1} + u_{2} + ... + u_{n-1} + u_{n}=\left (\dfrac{1}{2}+\dfrac{1}{3}+\ldots+\dfrac{1}{n+1} \right )-\left (\dfrac{1}{4}+\dfrac{1}{5}+\ldots+\dfrac{1}{n+3} \right )$
$S_n=\left (\dfrac{1}{2}+\dfrac{1}{3} \right )-\left (\dfrac{1}{n+2}+\dfrac{1}{n+3} \right )=\dfrac{n (13+5 n)}{6 (2+n) (3+n)}$
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