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

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

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

求证a1*a2*a3*a4.*an≥根号2n+1a1*a2*a3*a4.*an≥根号2n+1计算得a1=2,a2=4/3 ,a3=6/5.an=2n/2n-1
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)/(根号2n+1)≥根号(2(n+1)+1)
也就是证明(2n+2)(2n+2)≥(2n+1)(2(n+1)+1),该等式显然成立,证毕.

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