已知等差数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}\)满足\(S_{3}=0\)，\(S_{5}=-5\)，则数列\(\{ \dfrac {1}{a_{2n-1}a_{2n+1}}\}\)的前\(8\)项和为\((\)　　\()\) --优优班--学霸训练营

已知等差数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}\)满足\(S_{3}=0\)，\(S_{5}=-5\)，则数列\(\{ \dfrac {1}{a_{2n-1}a_{2n+1}}\}\)的前\(8\)项和为\((\)　　\()\)

A．\(- \dfrac {3}{4}\)

B．\(- \dfrac {8}{15}\)

C．\( \dfrac {3}{4}\)

D．\( \dfrac {8}{15}\)

【考点】等差数列的通项公式,数列的求和,裂项相消法,等差数列的求和

【分析】请登陆后查看

【解答】请登陆后查看

难度：中等

进入组卷