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            • 1.
              已知数列\(\{a_{n}\}\),其前\(n\)项和为\(S_{n}\),若函数\(y=x^{2}-2x\)在\(x=a_{n}\)处的切线斜率为\(S_{n}\),数列\(\{b_{n}\}\),满足点\((n,b_{n})(n∈N*)\)在直线\(y=x\)上.
              \((1)\)分别求\(\{a_{n}\}\),\(\{b_{n}\}\)的通项公式;
              \((2)\)求数列\(\{a_{n}b_{n}\}\)的前\(n\)项和\(T_{n}\).
            • 2.
              已知\(S_{n}\)为数列\(\{a_{n}\}\)的前\(n\)项和,且\(3S_{n}=1-a_{n}\).
              \((1)\)求数列\(\{a_{n}\}\)的通项公式;
              \((2)\)设\(b_{n}= \dfrac {1}{\log _{2}a_{n}\cdot \log _{2}a_{n+1}}\),求数列\(\{b_{n}\}\)的前\(n\)项和\(T_{n}\).
            • 3.
              已知等差数列\(\{a_{n}\}\)中,\(a_{3}=6\),\(a_{5}+a_{8}=26\).
              \((\)Ⅰ\()\)求数列\(\{a_{n}\}\)的通项公式;
              \((\)Ⅱ\()\)设\(b_{n}=2^{a_{n}}+n\),求数列\(\{b_{n}\}\)的前\(n\)项和\(S_{n}\).
            • 4.
              设\(S_{n}\)为数列\(\{a_{n}\}\)的前\(n\)项和,已知\(S_{n}=2a_{n}-2\).
              \((1)\)求数列\(\{a_{n}\}\)的通项公式;
              \((2)\)求数列\(\{ \dfrac {2-n}{a_{n}}\}\)的前\(n\)项和\(T_{n}\).
            • 5.
              已知数列\(\{a_{n}\}\)的前\(n\)项的和为\(S_{n}\),且\(S_{n}=2^{n}+n-1\),其中\(n∈N*\).
              \((\)Ⅰ\()\)求数列\(\{a_{n}\}\)的通项公式;
              \((\)Ⅱ\()\)若数列\(\{b_{n}\}\)满足\(b_{n}=2n(a_{n}-1)\),求数列\(\{b_{n}\}\)的前\(n\)项和\(T_{n}\).
            • 6.
              已知在\(\triangle ABC\)中,\(2B=A+C\),且\(c=2a\).
              \((1)\)求角\(A\),\(B\),\(C\)的大小;
              \((2)\)设数列\(\{a_{n}\}\)满足\(a_{n}=2^{n}|\cos nC|\),前\(n\)项和为\(S_{n}\),若\(S_{n}=20\),求\(n\)的值.
            • 7.
              已知数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),且\(2S_{n}=1-a_{n}(n∈N^{*}).\)
              \((\)Ⅰ\()\)求数列\(\{a_{n}\}\)的通项公式;
              \((\)Ⅱ\()\)设\(b_{n}= \dfrac {1}{\log _{ \frac {1}{3}}a_{n}}\),\(c_{n}= \dfrac { \sqrt {b_{n}b_{n+1}}}{ \sqrt {n+1}+ \sqrt {n}}\),求数列\(\{c_{n}\}\)的前\(n\)项和\(T_{n}\).
            • 8.
              已知\(n^{3}(n∈N^{*})\)有如下的拆分方式:\(1^{3}=1\),\(2^{3}=2+4+2\),\(3^{3}=3+6+9+6+3\),\(…\),这些通过拆分得到的数可组成右边的数阵:
              \((1)\)认真观察数阵,求和:\(1^{3}+2^{3}+…+n^{3}\);
              \((2)\)若数列\(\{a_{n}\}\)中的每一项都大于\(0\),证明:\(\{a_{n}\}\)的通项公式为\(a_{n}=n\)的充要条件是对任意的\(n∈N^{*}\),都有\( \dfrac { a_{ 1 }^{ 3 }+ a_{ 2 }^{ 3 }+…+ a_{ n }^{ 3 }}{a_{1}+a_{2}+…+a_{n}}= \dfrac {1}{2}n(n+1)\).
            • 9.
              已知数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=1-na_{n}(n∈N^{*})\)
              \((1)\)计算\(a_{1}\),\(a_{2}\),\(a_{3}\),\(a_{4}\);
              \((2)\)猜想\(a_{n}\)的表达式,并用数学归纳法证明你的结论.
            • 10.
              设\(S_{n}\)为各项不相等的等差数列\(\{a_{n}\}\)的前\(n\)项和,已知\(a_{3}a_{5}=3a_{7}\),\(S_{3}=9\).
              \((1)\)求数列\(\{a_{n}\}\)通项公式;
              \((2)\)设\(T_{n}\)为数列\(\{ \dfrac {1}{a_{n}a_{n+1}}\}\)的前\(n\)项和,求\( \dfrac {T_{n}}{a_{n+1}}\)的最大值.
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