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            • 1.

              用数学归纳法证明不等式\(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots +\dfrac{1}{2n-1} < n(n∈N*\),且\(n > 1)\)时,第一步应证明下述哪个不等式成立

              A.\(1 < 2\)
              B.\(1+\dfrac{1}{2} < 2\)
              C.\(1+\dfrac{1}{2}+\dfrac{1}{3} < 2\)
              D.\(1+\dfrac{1}{3} < 2\)
            • 2.

              当\(n\in {{N}^{*}}\)时,\({{S}_{n}}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\cdot \cdot \cdot +\dfrac{1}{2n-1}-\dfrac{1}{2n}\),\({{T}_{n}}=\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+\cdot \cdot \cdot +\dfrac{1}{2n}\),

              \((\)Ⅰ\()\)求\({{S}_{1}},{{S}_{2}},{{T}_{1}},{{T}_{2}}\);

              \((\)Ⅱ\()\)猜想\({{S}_{n}}\)与\({{T}_{n}}\)的关系,并用数学归纳法证明.

            • 3.

              当\(n∈N^{*}\)时,\({{S}_{n}}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots +\dfrac{1}{2n-1}\dfrac{1}{2n}{{T}_{n}}=\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+\cdots +\dfrac{1}{2n}\)

              \((1)\)求\(S_{1}\),\(S_{2}\),\(T_{1}\),\(T_{2}\).

              \((2)\)猜想\(S_{n}\)与\(T_{n}\)的关系,并用数学归纳法证明

            • 4.

              已知\(n∈N*\),求证:\(1·2^{2}-2·3^{2}+…+(2n-1)·(2n)^{2}-2n·(2n+1)^{2}=-n(n+1)(4n+3)\).

            • 5.

              函数\(f_{1}(x)=\dfrac{1}{x}\),\(f_{2}(x)=\dfrac{1}{x+{{f}_{1}}(x)}\),\(…\),\(f_{n+1}(x)=\dfrac{1}{x+{{f}_{n}}(x)}\),\(…\),则函数\(f_{2018}(x)\)是\((\)  \()\)

              A.奇函数但不是偶函数                    
              B.偶函数但不是奇函数

              C.既是奇函数又是偶函数                  
              D.既不是奇函数又不是偶函数
            • 6. 利用数学归纳法证明不等式\(1+ \dfrac {1}{2}+ \dfrac {1}{3}+ \dfrac {1}{4}+…+ \dfrac {1}{2^{n-1}+1} < f(n)(n\geqslant 2,n∈N^{*})\)的过程中,由\(n=k\)变到\(n=k+1\)时,左边增加了\((\)  \()\)
              A.\(1\)项
              B.\(k\)项
              C.\(2^{k-1}\)项
              D.\(2^{k}\)项
            • 7.

              用数学归纳法证明\(\dfrac{1}{n+1}+ \dfrac{1}{n+2}+⋯+ \dfrac{1}{3n}\geqslant \dfrac{5}{6} \)时,从\(n=k\)到\(n=k+1\),不等式左边需添加的项是\((\)    \()\)

              A.\(\dfrac{1}{3k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}\)
              B.\(\dfrac{1}{3k+1}+\dfrac{1}{3k+2}-\dfrac{2}{3k+3}\)
              C.\(\dfrac{1}{3k+3}-\dfrac{1}{k+1}\)
              D.\(\dfrac{1}{3k+3}\)
            • 8. 用数学归纳法证明等式:\(n∈N\),\(n\geqslant 1\),\(1- \dfrac {1}{2}+ \dfrac {1}{3}- \dfrac {1}{4}+\cdots + \dfrac {1}{2n-1}- \dfrac {1}{2n}= \dfrac {1}{n+1}+ \dfrac {1}{n+2}+\cdots + \dfrac {1}{2n}\).
            • 9.

              用数学归纳法证明“\((n+1)(n+2)\cdots (n+n)={{2}^{n}}\cdot 1\cdot 2\cdot \cdots \cdot (2n-1)\)”\((n\in {{N}_{+}})\)时,从“\(n=k到n=k+1 \)”时,左边的式子之比是(    )

              A.\(\dfrac{1}{2k+1}\)
              B.\(\dfrac{2k+3}{k+1}\)
              C.\(\dfrac{2k+1}{k+1}\)       
              D.\(\dfrac{1}{2(2k+1)}\)
            • 10.

              \((1)\)用数学归纳法证明\(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots \dfrac{1}{{{2}^{n}}-1} < n.(n\in {{N}^{*}},n > 1)\)时,第一步应验证的不等式是_______.

              \((2)\)已知\(i\)是虚数单位,计算\(\dfrac{(3-4i){{(1+i)}^{3}}}{4+3i}\)的结果为                      .

              \((3)\)若函数\(f\left( x \right)\)在定义域\(D\)内某区间\(I\)上是增函数,且\(\dfrac{f(x)}{x}\)在\(I\)上是减函数,则称\(y=f(x)\)在\(I\)上是“弱增函数”\(.\)已知函数\(h(x)={{x}^{2}}-(b-1)x+b\)在\((0,1]\)上是“弱增函数”,则实数\(b\)的值为_____________

              \((4)\)已知函数\(f(x)\)的定义域为\([-1,5]\),部分对应值如表,\(f(x)\)的导函数\(y=f{{'}}(x)\)的图象如图所示,


              \(x\)

              \(-1\)

              \(0\)

              \(4\)

              \(5\)

              \(f(x)\)

              \(1\)

              \(2\)

              \(2\)

              \(1\)

              下列关于\(f(x)\)的命题:

              \(①\)函数\(y=f(x)\)是周期函数;\(②\)函数\(y=f(x)\)在\([0,2]\)上减函数;

              \(③\)如果当\(x∈[-1,t]\)时,\(f(x)\)的最大值是\(2\),那么\(t\)的最大值是\(4\);

              \(④\)当\(1 < a < 2\)时,函数\(y=f(x)-a\)有\(4\)个零点;

              \(⑤\)函数\(y=f(x)-a\)的零点个数可能为\(0\),\(1\),\(2\),\(3\),\(4\).

              其中正确命题的序号是__________________\((\)写出所有正确命题的序号\()\).

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