知识点挑题 - 高中数学 - 优优班--学霸训练营

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5．

用数学归纳法证明“\(3^{4n+1}+5^{2n+1}\left( \left. n∈N \right. \right)\)能被\(8\)整除”时，当\(n=k+1\)时，对于\(3^{4(k+1)+1}+5^{2(k+1)+1}\)可变形为\((\)　　\()\)

A．\(56·3^{4k+1}+25\left( \left. 3^{4k+1}+5^{2k+1} \right. \right)\)

B．\(3^{4k+1}+5^{2k+1}\)

C．\(3^{4}×3^{4k+1}+5^{2}×5^{2k+1}\)

D．\(25\left( \left. 3^{4k+1}+5^{2k+1} \right. \right)\)

难度：较易

6．利用数学归纳法证明“\((n+1)(n+2)…(n+n)=2^{n}×1×3×…×(2n-1)\)，\(n∈N^{*}\)”时，从“\(n=k\)”变到“\(n=k+1\)”时，左边应增乘的因式是 ______ ．

难度：较易

解析收藏试题篮
7．已知\(\{a_{n}\}\)为等差数列，且\(a_{n}\neq 0\)，公差\(d\neq 0\)．
\((\)Ⅰ\()\)证明：\( \dfrac { C_{ 2 }^{ 0 }}{a_{1}}- \dfrac { C_{ 2 }^{ 1 }}{a_{2}}+ \dfrac { C_{ 2 }^{ 2 }}{a_{3}}= \dfrac {2d^{2}}{a_{1}a_{2}a_{3}}\)
\((\)Ⅱ\()\)根据下面几个等式：\( \dfrac {1}{a_{1}}- \dfrac {1}{a_{2}}= \dfrac {d}{a_{1}a_{2}}\)；\( \dfrac { C_{ 2 }^{ 0 }}{a_{1}}- \dfrac { C_{ 2 }^{ 1 }}{a_{2}}+ \dfrac { C_{ 2 }^{ 2 }}{a_{3}}= \dfrac {2d^{2}}{a_{1}a_{2}a_{3}}\)；\( \dfrac { C_{ 3 }^{ 0 }}{a_{1}}- \dfrac { C_{ 3 }^{ 1 }}{a_{2}}+ \dfrac { C_{ 3 }^{ 2 }}{a_{3}}- \dfrac { C_{ 3 }^{ 3 }}{a_{4}}= \dfrac {6d^{3}}{a_{1}a_{2}a_{3}a_{4}}\)

；\( \dfrac { C_{ 4 }^{ 0 }}{a_{1}}- \dfrac { C_{ 4 }^{ 1 }}{a_{2}}+ \dfrac { C_{ 4 }^{ 2 }}{a_{3}}- \dfrac { C_{ 4 }^{ 3 }}{a_{4}}+ \dfrac { C_{ 4 }^{ 4 }}{a_{5}}= \dfrac {24d^{4}}{a_{1}a_{2}a_{3}a_{4}a_{5}}\)，\(…\)
试归纳出更一般的结论，并用数学归纳法证明．

难度：较易

解析收藏试题篮

8．用数学归纳法证明：\(1+ \dfrac {1}{1+2}+ \dfrac {1}{1+2+3}+…+ \dfrac {1}{1+2+3+…+n}= \dfrac {2n}{n+1}\)时，由\(n=k\)到\(n=k+1\)左边需要添加的项是\((\)　　\()\)

A．\( \dfrac {1}{k(k+2)}\)

B．\( \dfrac {1}{k(k+1)}\)

C．\( \dfrac {1}{(k+1)(k+2)}\)

D．\( \dfrac {2}{(k+1)(k+2)}\)

难度：较易

9．用数学归纳法证明不等式\(1+ \dfrac {1}{2^{3}}+ \dfrac {1}{3^{3}}+…+ \dfrac {1}{n^{3}} < 2- \dfrac {1}{n}(n\geqslant 2,n∈N_{+})\)时，第一步应验证不等式\((\)　　\()\)

A．\(1+ \dfrac {1}{2^{3}} < 2- \dfrac {1}{2}\)

B．\(1+ \dfrac {1}{2^{3}}+ \dfrac {1}{3^{3}} < 2- \dfrac {1}{3}\)

C．\(1+ \dfrac {1}{2^{3}} < 2- \dfrac {1}{3}\)

D．\(1+ \dfrac {1}{2^{3}}+ \dfrac {1}{3^{3}} < 2- \dfrac {1}{4}\)

难度：较易

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