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            • 1.

              若数列\(\{a_{n}\}\)满足\({{a}_{11}}=\dfrac{1}{52}\),\(\dfrac{1}{{{a}_{n+1}}}-\dfrac{1}{{{a}_{n}}}=5(n\in {{N}^{*}})\),则\({{a}_{1}} = \)_______________ .

            • 2.

              已知等差数列\(\{a_{n}\}\)的公差为\(3\),若\(a_{1}\),\(a\),\(a_{4}\)成等比数列,则\(a_{2}=(\)    \()\)

              A.\(-9\)
              B.\(-6\)
              C.\(-8\)
              D.\(-10\)
            • 3.

              已知首项都是\(1\)的两个数列\(\left\{ {{a}_{n}} \right\},\left\{ {{b}_{n}} \right\}({{b}_{n}}\ne 0),n\in {{N}^{*}})\)满足\({{a}_{n}}{{b}_{n+1}}-{{a}_{n+1}}{{b}_{n}}+2{{b}_{n+1}}{{b}_{n}}=0.\)

              \((1)\)令\({{c}_{n}}=\dfrac{{{a}_{n}}}{{{b}_{n}}},\)求数列\(\left\{ {{c}_{n}} \right\}\)的通项公式;

              \((2\)若\({{b}_{n}}={{3}^{{n}}},\)求数列\(\left\{ {{a}_{n}} \right\}\)的前\(n\)项和\({{S}_{{n}}}\) .

            • 4.

              在数列\(\{\)\(a_{n}\)\(\}\)中,\(a\)\({\,\!}_{1}=1\),\(3\)\(a_{n}a_{n}\)\({\,\!}_{-1}+\)\(a_{n}\)\(-\)\(a_{n}\)\({\,\!}_{-1}=0(\)\(n\)\(\geqslant 2\),\(n\)\(∈N^{*}).\)

              \((1)\)求证:数列\(\left\{\begin{matrix} \dfrac{1}{a_{n}}\end{matrix}\right\}\)是等差数列;

              \((2)\)求数列\(\{\)\(a_{n}\)\(\}\)的通项公式;

              \((3)\)求数列\(\{a_{n}a_{n+1}\}\)的前\(n\)项和\(T_{n}\)

            • 5. 下面给出了四个类比推理,结论正确的是
              \(①\)若\(a,b,c\in R\)则\((ab)c=a(bc)\);类比推出:若\(\overset{\to }{{a}}\,,\overset{\to }{{b}}\,,\overset{\to }{{c}}\,\)为三个向量则\((\overset{\to }{{a}}\,\cdot \overset{\to }{{b}}\,)\overset{\to }{{\cdot c}}\,=\overset{\to }{{a}}\,\cdot (\overset{\to }{{b}}\,\cdot \overset{\to }{{c}}\,)\) .
              \(②\)在正三角形\(ABC\)中,若\(D\)是边\(BC \)的中点,\(G \)是三角形\(ABC \)的重心,则\( \dfrac{AG}{GD}=2 \);类比推出:在棱长都相等的四面体\(ABCD\)中,若\(\triangle ABC\)的中心为\(M\),四面体内部一点\(O\)到四面体各面的距离都相等,则\( \dfrac{AO}{OM}=3 \).
              \(③a\),\(b\)为实数,若\({a}^{2}+{b}^{2}=0 \)则\(a=b=0 \);类比推出:\({z}_{1},{z}_{2} \)为复数,若\(z_{1}^{2}+z_{2}^{2}=0\)则\({{z}_{1}}={{z}_{2}}=0\) .
              \(④\) 若\(\{a_{n}\}\)是等差数列,对于\({b}_{n}= \dfrac{1}{n}({a}_{1}+{a}_{2}+⋯+{a}_{n}) \),则\(\{b_{n}\}\)也是等差数列;
              类比推出:若\(\{c_{n}\}\)是各项都为正数的等比数列,\({{d}_{n}}=\sqrt[n]{{{c}_{1}}\cdot {{c}_{2}}\cdot {{c}_{3}}\cdot \cdots \cdot {{c}_{n}}}\),则\(\{d_{n}\}\)也是等比数列.
              A.\(①②\)          
              B.\(②③\)               
              C.\(②④\)             
              D.\(③④\)
            • 6.

              \((1)\)设向量\(\overset{\to }{{{a}}}\,\) \(=(-1,3)\),\(\overset{\to }{{{b}}}\,\) \(=(2,\)\(x\)\()\),若\(\overset{\to }{{{a}}}\,\) \(/\!/\)\(\overset{\to }{{{b}}}\,\) ,则\(x\)\(= \)______.


              \((2)\)已知\(p\)\(x\)\( < -3\)或\(x\)\( > 1\),\(q\)\(x\)\( > \)\(a\),若\(¬\)\(p\)是\(¬\)\(q\)的充分不必要条件,则\(a\)的取值范围______.



              \((3)\)若函数\(f(x)=\dfrac{4x}{x+4}\)  且\({{x}_{1}}=1\)  ,\({{x}_{n+1}}=f({{x}_{n}})\) ,则\({{x}_{2017}}\) \(= \)______.


              \((4)\)给出下列命题:
              \(①\)函数 \(y\)\(=\)\({\sin }(\dfrac{5\pi }{2}{-}2x)\) 是偶函数;
              \(②\)方程\(x=\dfrac{\pi }{8}\) 是函数 \(y\)\(=\) \(\sin \)\((2 \)\(x\)\(+ \dfrac{5π}{4} )\)的图象的一条对称轴方程;
              \(③\)若\(α\)、\(β\)是第一象限角,且\(α > β\),则 \(\sin \)\(α > \) \(\sin \)\(β\);
              \(④\)设 \(x\)\({\,\!}_{1}\)、 \(x\)\({\,\!}_{2}\)是关于 \(x\)的方程\(|\) \(\log _{a}x\)\(|=\) \(k\)\(( \)\(a\)\( > 0\), \(a\)\(\neq 1\), \(k\)\( > 0)\)的两根,则 \(x\)\({\,\!}_{1}\) \(x\)\({\,\!}_{2}=1\);
              其中正确命题的序号是______\(.(\)填出所有正确命题的序号\()\)
            • 7.

              设数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),命题\(P\):若\(S_{n}=2n^{2}+3n+1(n∈N^{*})\),则\(\{a_{n}\}\)为等差数列;命题\(q\):若\(S_{n}=2a_{n}+1(n∈N^{*})\),则\(\{a_{n}\}\)为等比数列,则下列说法正确的是\((\)    \()\)

              A.\(p\vee q\)为假
              B.\(p∧q\)为真
              C.\((\neg p)\wedge q\)为真
              D.\((\neg p)\vee q\)为假
            • 8.

              计算\(\sin {{15}^{0}}{{=}_{--------------}}\)

              已知数列\(\{\)\(a_{n}\)\(\}\)的前\(n\)项和\(S_{n}\)\(=\)\(n\)\({\,\!}^{2}+1\),则\(a_{n}\)\(=\)________

              一艘海轮从\(A\)处出发,以每小时\(40\)海里的速度沿南偏东\(40^{\circ}\)的方向直线航行,\(30\)分钟后到达\(B\)处,在\(C\)处有一座灯塔,海轮在\(A\)处观察灯塔,其方向是南偏东\(70^{\circ}\),在\(B\)处观察灯塔,其方向是北偏东\(65^{\circ}\),那么\(B\)\(C\)两点间的距离是________海里\(.\)                                               




              等差数列\(\{\)\(a_{n}\)\(\}\)的前\(n\)项和为\(S_{n}\),已知\(S\)\({\,\!}_{10}=0\),\(S\)\({\,\!}_{15}=25\),则\(nS_{n}\)\(+2017\)的最小值为________.

            • 9.

              数列\(\{{{a}_{n}}\}\)中,\({{a}_{1}}=1,{{a}_{n+1}}\cdot {{a}_{n}}+{{a}_{n+1}}-{{a}_{n}}=0,\)设\({{b}_{n}}=\dfrac{1}{{{a}_{n}}}\),

               \((1)\)证明数列\(\{{{b}_{n}}\}\)是等差数列,并求数列\(\{{{a}_{n}}\}\)的通项公式;

               \((2)\)设\({{c}_{n}}=[\dfrac{2{{b}_{n}}+3}{5}],\)求数列\(\{{{c}_{n}}\}\)的前\(8\)项和\({{S}_{n}}\),其中\([x]\)表示不超过\(x\)的最大整数,如\([0.9]=0,[2.6]=2\)。

            • 10.

              己知数列\(\{a_{n}\}\),\(a_{1}=1\),\(a_{n+1} > a_{n}\),\((a_{n}+a_{n+1}-1)^{2}=4a_{n}a_{n+1}(n∈N^{*}).\)

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

              \((2)\)记\({{b}_{n}}=\dfrac{1}{\sqrt[4]{{{a}_{n}}}}\),\(T_{n}=b_{1}+b_{2}+…+b_{n}\),估算\(T_{2017}\)的整数部分.

              \((\)参考数据:\(1.41 < \sqrt{2} < 1.42\),\(44.92 < \sqrt{2018} < 45)\)

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