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a(n)=2[s(n)]^(1/2)-1,a(1)=2[s(1)]^(1/2)-1=2[a(1)]^(1/2)-1,0={[a(1)]^(1/2)-1}^2,1=[a(1)]^(1/2), 1=a(1)=s(1).a(n+1)=s(n+1)-s(n)=2[s(n+1)]^(1/2)-1,{[s(n+1)]^(1/2)-1}^2=s(n)因a(n)>0,故s(n)>0.又,a(1)=1,所以, s(n)>=s(1)=a(1)=1. [s(n)]^(1/2) - 1 >=0.[s(n+1)]^(1/2) - 1 = [s(n)]^(1/2),[s(n+1)]^(1/2) - [s(n)]^(1/2) = 1.{[s(... |
