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            • 1.

              设各项均为正数的数列\(\left\{ {{a}_{n}} \right\}\)的前\(n\)项和为\({{S}_{n}}\),满足\(4{{S}_{n}}=a_{^{_{n+1}}}^{2}-4n-1\),且\({{a}_{1}}=1\),公比大于\(1\)的等比数列\(\left\{ {{b}_{n}} \right\}\)满足\({{b}_{2}}=3\),\({{b}_{1}}+{{b}_{3}}=10\).

              \((1)\)求证数列\(\left\{ {{a}_{n}} \right\}\)是等差数列,并求其通项公式;

              \((2)\)若\({{c}_{n}}=\dfrac{{{a}_{n}}}{3{{b}_{n}}}\),求数列\(\left\{ {{c}_{n}} \right\}\)的前\(n\)项和\({{T}_{n}}\);

              \((3)\)在\((2)\)的条件下,若\({{c}_{n}}\leqslant {{t}^{2}}+\dfrac{4}{3}t-2\)对一切正整数\(n\)恒成立,求实数\(t\)的取值范围.

            • 2.
              已知等差数列\(\{a_{n}\}\)的公差\(d\neq 0\),它的前\(n\)项和为\(S_{n}\),若\(S_{5}=70\),且\(a_{2}\),\(a_{7}\),\(a_{22}\)成等比数列.
              \((1)\)求数列\(\{a_{n}\}\)的通项公式;

              \((2)\)设数列\(\left\{ \dfrac{1}{{S}_{n}}\right\} \)的前\(n\)项和为\(T_{n}\),求证:\(T_{n} < \dfrac{3}{8} \).

            • 3.

              若各项均为正数的数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),且\(2 \sqrt[]{S_{n}}=a_{n}+1 (n∈N*)\).

              \((1)\)求数列\(\{a_{n}\}\)的通项公式;

              \((2)\)若正项等比数列\(\{b_{n}\}\),满足\(b_{2}=2\),\(2b_{7}+b_{8}=b_{9}\),求\(T_{n}=a_{1}b_{1}+a_{2}b_{2}+…+a_{n}b_{n}\);

              \((3)\)对于\((2)\)中的\(T_{n}\),若对任意的\(n∈N^{*}\),不等式\(λ·(-1)^{n} < \dfrac{1}{2^{n+1}}(T_{n}+21)\)恒成立,求实数\(λ\)的取值范围.

            • 4.

              设数列\(\{a_{n}\}\)为等差数列,数列\(\{b_{n}\}\)为等比数列\(.\)若\(a_{1} < a_{2}\),\(b_{1} < b_{2}\),且\({b}_{i}={{a}_{i}}^{2}(i=1,2,3) \),则数列\(\{b_{n}\}\)的公比为________.

            • 5. 已知数列\(\{a_{n}\}\)满足\(a_{1}=1\),\(a_{n}= \dfrac {a_{n-1}}{2a_{n-1}+1}(n∈N^{*},n\geqslant 2)\),数列\(\{b_{n}\}\)满足关系式\(b_{n}= \dfrac {1}{a_{n}}(n∈N^{*}).\)
              \((1)\)求证:数列\(\{b_{n}\}\)为等差数列;
              \((2)\)求数列\(\{a_{n}\}\)的通项公式.
            • 6.

              已知\(A\)、\(B\)、\(C\)为\(\triangle ABC\)的三个内角,向量\(m\)满足\(|m|=\dfrac{\sqrt{6}}{2}\),且\(m=(\sqrt{2}\sin \dfrac{B+C}{2},\cos \dfrac{B-C}{2})\),若\(A\)最大时,动点\(P\)使得\(|\overrightarrow{PB}|\)、\(|\overrightarrow{BC}|\)、\(|\overrightarrow{PC}|\)成等差数列,则\(\dfrac{|\overrightarrow{PA}|}{|\overrightarrow{BC}|}\)的最大值是

              A.\(\dfrac{2\sqrt{3}}{3}\)
              B.\(\dfrac{2\sqrt{2}}{3}\)
              C.\(\dfrac{\sqrt{2}}{4}\)
              D.\(\dfrac{3\sqrt{2}}{4}\)
            • 7.

              在等差数列\(\left\{{a}_{n}\right\} \)中,\({a}_{1}=-2008 \),其前\(n\)项和为\(S_{n}\),若\(\dfrac{{S}_{12}}{12}- \dfrac{{S}_{10}}{10}=2 \),则\(S_{2008}\)的值等于______ .

            • 8.

              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\(S_{n}\).

            • 9. 若数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),且满足\(a_{n}+2S_{n}S_{n-1}=0(n\geqslant 2)\),\(a_{1}= \dfrac{1}{2}\).
              \((1)\)求证:\(\left\{ \left. \dfrac{1}{S_{n}} \right. \right\}\)成等差数列;

              \((2)\)求数列\(\{a\)\({\,\!}_{n}\)\(\}\)的通项公式.

            • 10. 若数列\(\{a_{n}\}\)满足条件:存在正整数\(k\),使得\(a_{n+k}+a_{n-k}=2a_{n}\)对一切\(n∈N^{*}\),\(n > k\)都成立,则称数列\(\{a_{n}\}\)为\(k\)级等差数列.
              \((1)\)已知数列\(\{a_{n}\}\)为\(2\)级等差数列,且前四项分别为\(2\),\(0\),\(4\),\(3\),求\(a_{8}+a_{9}\)的值;
              \((2)\)若\(a_{n}=2n+\sin ωn(ω\)为常数\()\),且\(\{a_{n}\}\)是\(3\)级等差数列,求\(ω\)所有可能值的集合,并求\(ω\)取最小正值时数列\(\{a_{n}\}\)的前\(3n\)项和\(S_{3n}\);
              \((3)\)若\(\{a_{n}\}\)既是\(2\)级等差数列\(\{a_{n}\}\),也是\(3\)级等差数列,证明:\(\{a_{n}\}\)是等差数列.
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