§4 数学归纳法
课后训练案巩固提升
A组
1.如果f(n)=1+1/2+1/3+...+1/(3n"-" 1)(n∈N+),那么f(n+1)-f(n)等于( )
A.1/(3n+2) B.1/3n+1/(3n+1)
C.1/(3n+1)+1/(3n+2) D.1/3n+1/(3n+1)+1/(3n+2)
解析:∵f(n+1)=1+1/2+1/3+...+1/(3n"-" 1)+1/3n+1/(3"(" n+1")-" 2)+1/(3"(" n+1")-" 1),
f(n)=1+1/2+1/3+...+1/(3n"-" 1),
∴f(n+1)-f(n)=1/3n+1/(3"(" n+1")-" 2)+1/(3"(" n+1")-" 1)
=1/3n+1/(3n+1)+1/(3n+2).
答案:D
2.观察下列式子:1+1/2^2 <3/2,1+1/2^2 +1/3^2 <5/3,1+1/2^2 +1/3^2 +1/4^2 <7/4,...,则可归纳出1+1/2^2 +1/3^2 +...+1/("(" n+1")" ^2 )小于0( )
A.n/(n+1) B.(2n"-" 1)/(n+1) C.(2n+1)/(n+1) D.2n/(n+1)
解析:所猜测的分式的分母为n+1,而分子3,5,7,...,恰好是第(n+1)个正奇数,即2n+1.
答案:C
3.设f(x)是定义在正整数集上的函数,且f(x)满足"当f(k)≥k2成立时总可推出f(k+1)≥(k+1)2成立."则下列命题总成立的是( )
A.若f(3)≥9成立,则当k≥1时,均有f(k)≥k2成立
B.若f(5)≥25成立,则当k≤5时,均有f(k)≥k2成立
C.若f(7)<49成立,则当k≥8时,均有f(k) D.若f(4)=25成立,则当k≥4时,均有f(k)≥k2成立 解析:由数学归纳法原理可得,