求证a1*a2*a3*a4.*an≥根号2n+1

题目

求证a1*a2*a3*a4.*an≥根号2n+1
a1*a2*a3*a4.*an≥根号2n+1
计算得a1=2,a2=4/3 ,a3=6/5.an=2n/2n-1

答案

用数学归纳法证明证明：n=1时,显然有a1=2≥根号3成立;假设结论成立,则n+1时,有a(n+1)= 2(n+1)/(2(n+1)-1)=(2n+2)/(2n+1)从而有1*a2*a3*a4.*an*a(n+1)≥(根号2n+1)*(2n+2)/(2n+1)=(2n+2)/(根号2n+1)即只需证明(2n+2)/...