(1+b1)(1+b3)(1+b5).(1+b2n-1)>根号2n+1.bn=1/n请用数学归纳法证明..

Question

(1+b1)(1+b3)(1+b5).(1+b2n-1)>根号2n+1.bn=1/n
请用数学归纳法证明..

Answer 1

n=1时,1+b1=1+1/1=2>√3=√(2×1+1)成立 假设n=k时命题成立,即(1+b1)(1+b3)...(1+b2k-1)>√(2k+1) n=k+1时,(1+b1)(1+b3)...(1+b2k-1)(1+b2k+1)>√(2k+1)×[1+1/(2k+1)]=√(2k+1)×(2k+2)/(2k+1) =(2k+2)/√(2k+1)=√[(2k+2)/(2k+1)]=√[(4k+8k+4)/(2k+1)]>√[(4k+8k+3)/(2k+1)] =√[(2k+1)(2k+3)/(2k+1)]=√(2k+3),即n=k+1时命题也成立 综上,原命题得证