Question

用数学归纳法证明1+n/2<=1+1/2+1/3+...+1/(2^n)<=1/2+n

Answer 1

f(n)=1+1/2+1/3+...+1/(2^n)-1-n/2
g(n)=1+1/2+1/3+...+1/(2^n)-1/2-n

f(1)=1+1/2-1-1/2=0
若f(n)≥0
f(n+1)=1+1/2+1/3+...+1/(2^n)-1-n/2+1+n/2-1-(n+1)/2+1/(2^n +1)+…1/2^(n +1)
而f(n)≥0
1/(2^n +1)+…1/2^(n +1)
≥[2^(n+1)-2^n-1+1]/2^(n+1)=1/2
f(n+1)≥0

同理：g(n)≤0

Answer 2

放大法，缩小法
分母放大缩小

Answer 3

n=1时,2=2成立
假设n=k时,(k+1)(k+2)(k+3).(k+k)=(2^k)*1*3*.(2k-1)成立
则当n=k+1时,
(k+2)(k+3).(k+1+k-1)(k+1+k)(k+1+k+1)
=(k+2)(k+3).(k+k)(k+1+k)2(k+1)
=(2^k)*1*3*.(2k-1)*2*(2k+1)
=(2^k+1)*1*3*.(2k-1)(2k+1)
所以：(n+1)(n+2)(n+3).(n+n)=(2^n)*1*3*.(2n-1)
好辛苦
给分吧