(1):因为数列{an}为等差数列,且a1=1,则由等差数列性质可得:前n项和Sn=a1n-(n(n-1)/2)*D即Sn=n-(n(n-1)/2)*D ,S2n=2n-(2n(2n-1)/2)*D 且 S2n/Sn=(4n+2)/(n+1),n=1,2,3``````.(1),则将Sn,S2n代入(1)式,化简可得(2)式....
在等差数列{an}中,a1=1,前n项的和sn满足条件S2n/S2=(4n+2)/(n+1),n=1,2.
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