1/1.2.3 + 1/2.3.4 + 1/3.4.5 +……+ 1/A.(A+1)(A+2) 04/09/2021 Bởi Adeline 1/1.2.3 + 1/2.3.4 + 1/3.4.5 +……+ 1/A.(A+1)(A+2)
Đáp án: $A=\dfrac{1}{2}(\dfrac{1}{1.2}-\dfrac{1}{(a+1)(a+2)})$ Giải thích các bước giải: $A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+…+\dfrac{1}{a(a+1)(a+2)}\\ \text{ta có:} \\ \dfrac{1}{n.(n+1)(n+2)}=\dfrac{n+2-n}{2n(n+1)(n+2)}=\dfrac{1}{2}(\dfrac{1}{n.(n+1)}-\dfrac{1}{(n+1)(n+2)})\\ \rightarrow A=\dfrac{1}{2}(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+…+\dfrac{1}{a.(a+1)}-\dfrac{1}{(a+1)(a+2)})\\ \rightarrow A=\dfrac{1}{2}(\dfrac{1}{1.2}-\dfrac{1}{(a+1)(a+2)})$ Bình luận
Đáp án:
$A=\dfrac{1}{2}(\dfrac{1}{1.2}-\dfrac{1}{(a+1)(a+2)})$
Giải thích các bước giải:
$A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+…+\dfrac{1}{a(a+1)(a+2)}\\ \text{ta có:} \\ \dfrac{1}{n.(n+1)(n+2)}=\dfrac{n+2-n}{2n(n+1)(n+2)}=\dfrac{1}{2}(\dfrac{1}{n.(n+1)}-\dfrac{1}{(n+1)(n+2)})\\ \rightarrow A=\dfrac{1}{2}(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+…+\dfrac{1}{a.(a+1)}-\dfrac{1}{(a+1)(a+2)})\\ \rightarrow A=\dfrac{1}{2}(\dfrac{1}{1.2}-\dfrac{1}{(a+1)(a+2)})$