Đáp án:
$A=\dfrac{1}{2}(\dfrac{1}{1.2}-\dfrac{1}{(a+1)(a+2)})$
Giải thích các bước giải:
$A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{a(a+1)(a+2)}\\ \text{ta có:} \\ \dfrac{1}{n.(n+1)(n+2)}=\dfrac{n+2-n}{2n(n+1)(n+2)}=\dfrac{1}{2}(\dfrac{1}{n.(n+1)}-\dfrac{1}{(n+1)(n+2)})\\ \rightarrow A=\dfrac{1}{2}(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{a.(a+1)}-\dfrac{1}{(a+1)(a+2)})\\ \rightarrow A=\dfrac{1}{2}(\dfrac{1}{1.2}-\dfrac{1}{(a+1)(a+2)})$
