优优班--学霸训练营 > 知识点挑题
全部资源
          排序:
          最新 浏览

          50条信息

            • 1.

              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.\(\dfrac{1}{3k+3}\)
            • 2.

              设\(f(k)\)是定义在正整数集上的函数,满足“只要\(f(k)\geqslant k^{2}\)成立就能推出\(f(k+1)\geqslant (k+1)^{2}\)成立”,则下列命题总成立的是  \((\)    \()\)

              A.若\(f(1) < 1\)成立,则\(f(10) < 100\)成立
              B.若\(f(2) < 4\)成立,则\(f(1)\geqslant 1\)成立
              C.若\(f(3)\geqslant 9\)成立,则当\(k\geqslant 1\)时,均有\(f(k)\geqslant k^{2}\)成立
              D.若\(f(4)\geqslant 25\)成立,则当\(k\geqslant 4\)肘,均有\(f(k)\geqslant k^{2}\)成立
            • 3.

              用数学归纳法证明“当\(n\)为正奇数时,\(x^{n}+y^{n}\)能被\(x+y\)整除”,第二步归纳假设应写成\((\)  \()\)

              A.假设\(n=2k+1(k∈N^{*})\)正确,再推\(n=2k+3\)正确
              B.假设\(n=2k-1(k∈N^{*})\)正确,再推\(n=2k+1\)正确
              C.假设\(n=k(k∈N^{*})\)正确,再推\(n=k+1\)正确

              D.假设\(n=k(k\geqslant 1)\)正确,再推\(n=k+2\)正确
            • 4.

              用数学归纳法证明等式\((n+1)(n+2)…(n+n)=2^{n}·1·3·…·(2n-1)\),第二步从\(n=k\)时到\(n=k+1\)时左端应增乘的代数式是  \((\)    \()\)

              A.\(2k+1\)
              B.\(2(2k+1)\)
              C.\(\dfrac{{2}k+{1}}{k+{1}}\)
              D.\(\dfrac{{2}k+{3}}{k+{1}}\)
            • 5.

              用数学归纳法证明\(1+2+3+\cdots +{{n}^{2}}=\dfrac{{{n}^{4}}+{{n}^{2}}}{2},\)则当\(n=k+1\)时左端应在\(n=k\)的基础上加上

              A.\({{k}^{2}}+1\)
              B.\({{(k+1)}^{2}}\)
              C.\(\dfrac{{{(k+1)}^{4}}+{{(k+1)}^{2}}}{2}\)
              D.\(({{k}^{2}}+1)+({{k}^{2}}+2)+\cdots +{{(k+1)}^{2}}\)
            • 6.
              记\([x]\)为不超过实数\(x\)的最大整数,例如:\([2]=2\),\([1.5]=1\),\([-0.3]=-1\),设\(a\)为正整数,数列\(\{x_{n}\}\)满足:\(x_{1}=a\),\(x_{n+1}=[ \dfrac {x_{n}+[ \dfrac {a}{x_{n}}]}{2}](n∈N^{*})\),现有下列命题:
              \(①\)当\(a=5\)时,数列\(\{x_{n}\}\)的前\(3\)项依次为\(5\),\(3\),\(2\);
              \(②\)对数列\(\{x_{n}\}\)都存在正整数\(k\),当\(n\geqslant k\)时,总有\(x_{n}=x_{k}\);
              \(③\)当\(n\geqslant 1\)时,\(x_{n} > \sqrt {a}-1\);
              \(④\)对某个正整数\(k\),若\(x_{k+1}\geqslant x_{k}\),则\(x_{n}=[ \sqrt {a}]\);
              其中的真命题个数为\((\)  \()\)
              A.\(4\)
              B.\(3\)
              C.\(2\)
              D.\(1\)
            • 7.
              用数学归纳法证明不等式“\(1+ \dfrac {1}{2}+ \dfrac {1}{3}+…+ \dfrac {1}{2^{n}-1} < n(n∈N^{*},n\geqslant 2)\)”时,由\(n=k(k\geqslant 2)\)不等式成立,推证\(n=k+1\)时,左边应增加的项数是\((\)  \()\)
              A.\(2^{k-1}\)
              B.\(2^{k}-1\)
              C.\(2^{k}\)
              D.\(2^{k}+1\)
            • 8.
              用数归纳法证明当\(n\)为正奇数时,\(x^{n}+y^{n}\)能被\(x+y\)整除,\(k∈N^{*}\)第二步是\((\)  \()\)
              A.设\(n=2k+1\)时正确,再推\(n=2k+3\)正确
              B.设\(n=2k-1\)时正确,再推\(n=2k+1\)时正确
              C.设\(n=k\)时正确,再推\(n=k+2\)时正确
              D.设\(n\leqslant k(k\geqslant 1)\)正确,再推\(n=k+2\)时正确
            • 9.
              用数学归纳法证明\(1+ \dfrac {1}{2}+ \dfrac {1}{3}+…+ \dfrac {1}{2^{n}-1} < n(n∈N^{*},n > 1)\)时,第一步应验证不等式\((\)  \()\)
              A.\(1+ \dfrac {1}{2} < 2\)
              B.\(1+ \dfrac {1}{2}+ \dfrac {1}{3} < 3\)
              C.\(1+ \dfrac {1}{2}+ \dfrac {1}{3}+ \dfrac {1}{4} < 3\)
              D.\(1+ \dfrac {1}{2}+ \dfrac {1}{3} < 2\)
            • 10.

              下列说法中正确的有(    )个

              \((1)\)已知命题\(p\):\(∃{x}_{φ}∈R,使{2}^{{x}_{φ}}=4 \),则\(\neg p\)是真命题

              \((2)\)等差数列\(\{a_{n}\}\)的公差为\(d\),前\(n\)项和为\(S_{n}\),则“\(d > 0\)”是“\(S_{4}+S_{6} > 2S_{5}\)”的充分必要条件

              \((3)\)用反证法证明“若\(a+b+c < 3\),则\(a\),\(b\),\(c\)中至少有一个小于\(1\)”时,“假设”应为“假设\(a\),\(b\),\(c\)全部都大于\(1\)”

              \((4)\)用数学归纳法证明\(1+ \dfrac{1}{2}+ \dfrac{1}{3}+…+ \dfrac{1}{{2}^{n}-1} < n (n∈N\)且\(n > 1)\),第二步证明中从“\(k\)到\(k+1\)”时,左端增加的项数是\(2^{k}-1\) 

              A.\(4\)个          
              B.\(3\)个           
              C.\(2\)个  
              D.\(1\)个
            0/40

            进入组卷