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            • 1.
              等比数列\(\{a_{n}\}\)的前\(n\)项和记为\(S_{n}\),若\( \dfrac {S_{2n}}{S_{n}}=3\),则\( \dfrac {S_{3n}}{S_{2n}}=\) ______ .
            • 2.
              已知数列\(\{\{a_{n}\}\)满足\(a_{1}=1,a_{n+1}= \dfrac {a_{n}}{a_{n}+2}\),\(b_{n+1}=(n-λ)( \dfrac {1}{a_{n}}+1)(n∈N^{*}),b_{1}=-λ\).
              \((1)\)求证:数列\(\{ \dfrac {1}{a_{n}}+1\}\)是等比数列;
              \((2)\)若数列\(\{b_{n}\}\)是单调递增数列,求实数\(λ\)的取值范围.
            • 3.
              数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),已知\(a_{1}= \dfrac {1}{5}\),且对任意正整数\(m\),\(n\),都有\(a_{m+n}=a_{m}a_{n}\),若对任意\(n∈N^{*}\),\(S_{n} < t\)恒成立,则实数\(t\)的取值范围是 ______ .
            • 4.
              设等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),则\(S_{4}\),\(S_{8}-S_{4}\),\(S_{12}-S_{8}\),\(S_{16}-S_{12}\)成等差数列\(.\)类比以上结论有:设等比数列\(\{b_{n}\}\)的前\(n\)项积为\(T_{n}\),则\(T_{4}\), ______ , ______ ,\( \dfrac {T_{16}}{T_{12}}\)成等比数列.
            • 5.
              设等比数列\(\{a_{n}\}\)满足\(a_{1}=1\),\(a_{3}+a_{5}=6\),则\(a_{5}+a_{7}+a_{9}=\) ______ .
            • 6.
              已知等差数列\(\{a_{n}\}\)的公差为\(2\),若\(a_{1}\),\(a_{3}\),\(a_{4}\)成等比数列,则\(a_{2}=\) ______ .
            • 7.
              各项为正数的等比数列\(\{a_{n}\}\),\(a_{4}⋅a_{7}=8\),则\(\log _{2}a_{1}+\log _{2}a_{2}+…+\log _{2}a_{10}=(\)  \()\)
              A.\(5\)
              B.\(10\)
              C.\(15\)
              D.\(20\)
            • 8.
              若数列\(\{a_{n}\}\)满足\(a_{1}=-1,a_{n}=2a_{n-1}-1(n∈N^{*},n\geqslant 2)\).
              \((1)\)求证:数列\(\{a_{n}-1\}\)是等比数列,并求数列\(\{a_{n}\}\)的通项公式;
              \((2)\)设\(b_{n}=\log _{2}(1-a_{n})\),若数列\(\{ \dfrac {1}{b_{n+1}b_{n}}\}(n∈N^{*})\)的前\(n\)项和为\(T_{n}\),求证:\(T_{n} < 1\).
            • 9.

              数列\({ }\!\!\{\!\!{ }{{a}_{n}}{ }\!\!\}\!\!{ }\)是公比为\(2\)的等比数列,其前\(n\)项和为\({{S}_{n}}.\)若\({{a}_{2}}=\dfrac{1}{2}\),则\({{a}_{n}}=\)____;\({{S}_{5}}=\)____.

            • 10.
              等比数列\(\{a_{n}\}\)的前三项和\(S_{3}=13\),若\(a_{1}\),\(a_{2}+2\),\(a_{3}\)成等比数列,则公比\(q=(\)  \()\)
              A.\(3\)或\( \dfrac {1}{3}\)
              B.\(-3\)或\( \dfrac {1}{3}\)
              C.\(3\)或\(- \dfrac {1}{3}\)
              D.\(-3\)或\(- \dfrac {1}{3}\)
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