设an=根号下n(n+1) 数列an前n项和为sn ,求证:[n(n+1)]/2
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设an=根号下n(n+1) 数列an前n项和为sn ,求证:[n(n+1)]/2
an=[n(n+1)]^(1/2)=0,故n+1>=(n+3)/2
所以Sn=n(n+3)/2(n*n)^(1/2)=n
Sn>1+2+……+n=n(n+1)/2
所以Sn=n(n+3)/2(n*n)^(1/2)=n
Sn>1+2+……+n=n(n+1)/2
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