数列\(\{a_{n}\}\)满足\(a_{1}=1\)，\({{a}_{n}}+\dfrac{{{a}_{n+1}}}{2{{a}_{n+1}}-1}=0\) ． \((\)Ⅰ\()\)求证：数列\(\left\{ \dfrac{1}{{{a}_{n}}} ight\}\) 是等差数列； \((\)Ⅱ\()\)若数列\(\{b_{n}\}\)满足\(b_{1}=2\)，\(\dfrac{{{b}_{n+1}}}{{{b}_{n}}}=\dfrac{2{{a}_{n}}}{{{a}_{n+1}}}\) ，求\(\{b_{n}\}\)的前\(n\)项和\(S

数列\(\{a_{n}\}\)满足\(a_{1}=1\)，_{\({{a}_{n}}+\dfrac{{{a}_{n+1}}}{2{{a}_{n+1}}-1}=0\)}．

\((\)Ⅰ\()\)求证：数列_{\(\left\{ \dfrac{1}{{{a}_{n}}} \right\}\)}是等差数列；

\((\)Ⅱ\()\)若数列\(\{b_{n}\}\)满足\(b_{1}=2\)，_{\(\dfrac{{{b}_{n+1}}}{{{b}_{n}}}=\dfrac{2{{a}_{n}}}{{{a}_{n+1}}}\)}，求\(\{b_{n}\}\)的前\(n\)项和\(S_{n}\)．

【考点】错位相减法,数列的求和,等差数列的判定与证明

【分析】请登陆后查看

【解答】请登陆后查看

难度：难

收藏试题篮