(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,故可取特值,将n=1代入(2)式,算出D=1则数列{an}的通项公式an=a1-(n-1)*D=1-n+1=n即:an=n
(n=1,2,3...) (为严密起见,可简单的用数学归纳法验证)(2):因为bn=anpan(p>0)...(3),则将an=n代入...
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