Question

用数学归纳法证明：3/2*5/4*7/6*9/8*...2n+1/2n> 根号下n+1.这道题怎么证明

Answer 1

证明：当n=1时

3/2>√2成立

假设当n=k时命题成立 即 3/2*5/4*7/6*9/8*...2k+1/2k>√(k+1)成立

即（3/2*5/4*7/6*9/8*...2k+1/2k）^2>k+1成立

即要证当 n=k+1时

3/2*5/4*7/6*9/8*...2k+1/2k*(2k+3)/(2k+2)>√(k+2)

两边平方

得到[3/2*5/4*7/6*9/8*...2k+1/2k*(2k+3)/(2k+2)]^2>(k+2)

因为（3/2*5/4*7/6*9/8*...2k+1/2k）^2>k+1

（3/2*5/4*7/6*9/8*...2k+1/2k）^2*[(2k+3)/(2k+2)]^2>(k+1)

[(1+1/2(k+1)]^2=(k+1)[1+1/(k+1)+1/4(k+1)^2]=k+2+1/4(k+1)^2>k+2

n=k+1时命题成立

得证