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            • 1.
              已知数列\(\{a_{n}\}\)满足:\(a_{n+1}=2a_{n}-n+1(n∈N^{*})\),\(a_{1}=3\).
              \((1)\)证明数列\(b_{n}=a_{n}-n(n∈N^{*})\)是等比数列,并求数列\(\{a_{n}\}\)的通项;
              \((2)\)设\(c_{n}= \dfrac {a_{n+1}-a_{n}}{a_{n}a_{n+1}}\),数列\(\{c_{n}\}\)的前\(n\)项和为\(\{S_{n}\}\),求证:\(S_{n} < 1\).
            • 2.

              已知在\(\triangle ABC\)中,角\(A\),\(B\),\(C\)所对的边分别为\(a\),\(b\),\(c\),且满足\(\dfrac{\cos B}{b}+\dfrac{\cos C}{c}=\dfrac{\sin A}{2\sin C}\).

              \((1)\)试求\(b\)的值;

              \((2)\)若\(\sqrt{3}b\cos C=(2a-\sqrt{3}c)\cos B\),且\(\sin A\),\(\sin B\),\(\sin C\)成等比数列,试求\(\triangle ABC\)的面积.

            • 3.
              已知\(6\),\(a\),\(b\),\(48\)成等差数列,\(6\),\(c\),\(d\),\(48\)成等比数列,则\(a+b+c+d\)的值为 ______ .
            • 4.
              等比数列\(\{a_{n}\}\)的各项均为正数,且\(a_{1}+2a_{2}=4\),\(a_{4}^{2}=4a_{3}a_{7}\),则\(a_{5}=(\)  \()\)
              A.\( \dfrac {1}{8}\)
              B.\( \dfrac {1}{16}\)
              C.\(20\)
              D.\(40\)
            • 5.
              已知等比数列\(\{a_{n}\}\)的公比为\(- \dfrac {1}{2}\),则\( \dfrac {a_{1}+a_{3}+a_{5}}{a_{2}+a_{4}+a_{6}}\)的值是\((\)  \()\)
              A.\(-2\)
              B.\(- \dfrac {1}{2}\)
              C.\( \dfrac {1}{2}\)
              D.\(2\)
            • 6. 在数列{an}中,an+1=2an,若a5=4,则a4a5a6= ______
            • 7. 在数列\(\{a_{n}\}\)中,\(a_{n+1}=2a_{n}\),若\(a_{5}=4\),则\(a_{4}a_{5}a_{6}=\) ______ .
            • 8. 等比数列\(\{a_{n}\}\)的各项均为正数,且\(2a_{1}+3a_{2}=1\),\(a_{3}^{2}=9a_{2}a_{6}\),
              \((\)Ⅰ\()\)求数列\(\{a_{n}\}\)的通项公式;
              \((\)Ⅱ\()\)设\(b_{n}=\log _{3}a_{1}+\log _{3}a_{2}+…+\log _{3}a_{n}\),求数列\(\{ \dfrac {1}{b_{n}}\}\)的前\(n\)项和.
            • 9. 已知\(\{a_{n}\}\)是公比大于\(1\)的等比数列,若\(2a_{1}\),\( \dfrac {3}{2}a_{2}\),\(a_{3}\)成等差数列,则\( \dfrac {S_{4}}{a_{4}}=(\)  \()\)
              A.\( \dfrac {31}{16}\)
              B.\( \dfrac {15}{16}\)
              C.\( \dfrac {15}{8}\)
              D.\(2\)
            • 10.
              在数列\(\{a_{n}\}\)中,\(a_{n+1}=2a_{n}\),若\(a_{5}=4\),则\(a_{4}a_{5}a_{6}=\) ______ .
            0/40

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