已知等差数列前n项和为sn,s4=24,s7=63
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通项an=19+(n-1)*(-2)=21-2nSn=(a1+an)n/2=(19+21-2n)n/2=-n²+20n
Sn+1/(2n+1)-Sn/(2n-1)=1Sn/(2n-1)=S1+n-1→Sn=(S1+n-1)(2n-1)→Sn=n(2n-1)an=4n-31/√an=2/2√(4n-3)>2/(√4n-3
设:等差数列{an}的公差为d,通项为an=a1+(n-1)d,则:sn=a1+a2+...+an=na1+n(n-1)d/2lim(n->∞)(n*an)/Sn=lim(n->∞)[n*(a1+(n
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
唉,你太粗心了吧~我给你修正下(向我现在这样的好人不多了哈哈~!)Sm/Sn=(m^2)/(n^2),求am/an?对吧,很简单的呦am/an=2am/(2an)=a1+a2m-1/(a1+a2n-1
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
在{an}为等差数列中,当m+n=p+q(m,n,p,q∈N+)时,am+an=ap+aq.所以a2+a20b7+b15=21×(a1+a21)×1221×(b1+b21)×12=S21T21,又因为
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
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.
由题,a4+a5+a6+a7+a8=0所以a6=0,当n>7时,有:Sn-S7=a8+a9+……+an=0n=7显然成立n
∵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