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...答:(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.因为(1)式对任意正整数都成立, ...
在数列{An}中,A1=1,当n≥2时,其前项和Sn满足:2Sn²=An(2Sn-1)?答:∴1/Sn-1/S=2,∴数列{1/Sn}是等差数列,公差为2,∴1/Sn=1/S1+2(n-1)=2n-1.∴Sn=1/(2n-1).,6,在数列{An}中,A1=1,当n≥2时,其前项和Sn满足:2Sn²=An(2Sn-1)求证:数列{1/Sn}是等差数列,并用n表示Sn