由二项式定理,当a>0,k>1时,(1+a)^k = C(k,0)+C(k,1)*a+...+C(k,k)*a^k > C(k,0)+C(k,1)*a = 1+na∴(3/2)^(n-1) = (1+1/2)^(n-1) > 1+(n-1)/2 = (n+1)/2∴(3/2)^(n-1) > 2/(n+1)