已知等差数列an的前n项和为sn,s7＝63

通项an=19+(n-1)*(-2)=21-2nSn=(a1+an)n/2=(19+21-2n)n/2=-n²+20n

因为Sn-Sn-1=n^2-3n-{(n-1)^2-3(n-1)}=2n-4.又由an=Sn-Sn-1,所以an=2n-4,最后还要验证一下,当n=1时,S1=a1,符合题意.d=an-an-1=2易

S6=(a1+a6)*6/2=362a1+5d=12Sn-S(n-6)=180即[a(n-5)+an]*6/2=180最后6项的和是6an-15d=1802an-5d=60相加2(a1+an)=72S

（Ⅰ）∵等比数列{an}的前n项和为Sn，S1，S3，S2成等差数列，∴2（a1+a1q+a1q2）=a1+a1+a1q，解得q=-12或q=0（舍）．∴q=-12．（Ⅱ）∵a1-a3=3，q=-12

由题意可得a1b1=S1T1=524=13，故a1=13b1．设等差数列{an}和{bn}的公差分别为d1 和d2，由S2T2=a1+a1+d 1b1+b1 +d&nbs

1、a4-a1=-9=3dd=-3an=25-3(n-1)=-3n+28an>0-3n+28>0n0,a10S8S9>S10所以n=9.Sn最大2、a2=a1+d=22a20=-60+28=-32有1

因为Sn=324,s(n-6)=144所以最后六项和=324-144=180=a(n-5)+a(n-4)+,+an又S6=36=a1+a2+,+a6两侧同时相加,有6(a1+an)=216a1+an=

因为a1＝S1＝(a1+12)2，所以 a1=1．设公差为d，则有a1+a2=2+d=S2＝(2+d2)2．解得d=2或d=-2（舍）．所以an=2n-1，Sn＝n2．所以 bn＝

等差数列前n项和Sn=na1+n*（n-1）*d/2n=6时S6=6a1+6*5*d/2S6=6a1+15d36=6a1+15da1=6-(5/2)dSn=na1+n*(n-1)*d/2=324将a1

S(n)＝n^2－9nS(n-1)=(n-1)^2-9(n-1)=n^2-2n+1-9n+9=n^2-11n+10a(n)=S(n)-S(n-1)=(n^2－9n)-(n^2-11n+10)=2n-1

S1=a1S2=a1(1+q)S3=a1(1+q+q^2)S1,S3,S2成等差数列即s3-s1=s2-s31+q+q^2-1=1+q-(1+q+q^2)q^2+q=-q^2q=0或-1/2如果a1-

显然的有d060+12*7+42d>0即d>-24/7类似的有156+52d

令n=9，得到S9T9＝7×9+29+3=6512，又S9=9(a1+a9) 2=9a5，T9=9(b1+b9) 2=9b5，∴S9T9=9a59b5=a5b5=6512．故答案为

∵SnTn=2n3n+1，∴anbn=a1+a2n−1b1+b2n−1=S2n−1T2n−1=2(2n−1)3(2n−1)+1=2n−13n−1∴limn→∞anbn=limn→∞2n−13n−1=l

1.通项:an=19+(n-1)*(-2)=21-2nSn=(a1+an)n/2=(19+21-2n)n/2=-n²+20n2.bn-an=3^(n-1)bn=21-2n+3^(n-1){b

S12=6(a6+a7)>0a6+a7>0S13=13*a7-a7绝对值最小的是第7项

证明：设等差数列{an}的首项为a1，公差为d，则Sn=na1+n(n−1)d2．bn＝Snn＝a1+n−12d．则bn+1−bn＝a1+n2d−a1−n−12d＝d2．∴数列{bn}是等差数列．

∵等差数列{an}{bn}的前n项和分别为Sn，Tn，∵SnTn＝7nn+3，∴a5b5=s9T9＝7×99+3=6312＝214，故答案为：214

假设m>nSn=A1+A2+……+AnSm=A1+A2+……+An+A(n+1)+A(n+2)+……+AmSm-Sn=A(n+1)+A(n+2)+……+Am=0(共m-n项)从A(n+1)项到Am项也

当n=1时,a1=S1=1当n≥2时,an=Sn-S(n-1)=3n²-2n-3(n-1)²+2(n-1)=6n-5∵当n=1时,满足an=6n-5又∵an-a(n-1)=6n-5