已知等差数列an的前n项和Sn,且a3=7,s11=143
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Sm/Sn=m²/n²则:am/Sn=(Sm-Sm-1)/Sn=(m²-(m-1)²)/n²=(2m-1)/n²同理:an/Sn=(2n-1
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
(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)设等差数列{an}的公差为d,∵前n项和Sn满足S3=0,S5=-5,∴3a1+3d=05a1+10d=−5,解得a1=1,d=-1.∴an=1-(n-1)=2-n.(2)1a2n−1a2n+1
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
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-
a2=a1qa8=a1q^7a5=a1q^42a8=a2+a52a1q^7=a1q+a1q^42q^6=1+q^32q^6=1+q^32q^6-q^3-1=0(2q^3+1)(q^3-1)=0q^3=
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
S12>0,S1307d+24>0d>-24/7S13=(a1+a1+12d)*13/2=(2a1+12d)*13/2=13(a1+6d)=13(a1+2d+4d)=13(a3+4d)=13(12+4
等差数列求和公式:Sn=n*a1+n*(n-1)*d/2S12=12*a1+12*11*d/2=12*a1+66d>0得a1+5.5d>0S13=13*a1+13*12*s/2=13*a1+78d
n=1时,a1=S1=a+bn≥2时,Sn=a×n²+bnS(n-1)=a×(n-1)²+b两式相减得:an=Sn-S(n-1)=2a×n-a∴a(n-1)=2a×(n-1)-a∴
知道Sn,求an,需记住an=Sn-Sn-1当n=1是an=Sn=n²=1当n>=2时an=Sn-Sn-1=n²-(n-1)^2=2n-1a1=1也符合此式则an=2n-1再问:做
Sn=((An+1)/2)^2A1=S1=((A1+1)/2)^2(A1-1)^2=0A1=1Sn=n(A1+An)/2=n(1+An)/2=((An+1)/2)^2(An+1)/2=nAn=2n-1
设公差为dS12=(a3+a10)*6=(2a3+7d)*6=(24+7d)*6>0S13=a7*13=(a3+4d)*13=(12+4d)*130且12+4d
a3=a1*q^2;a9=a1*q^8;a6=a1*q^5;因为a3,a9,a6是等差数列,所以,2a9=a3+a6.化简,2q^9=q^3+q^6.s3+s6=a1*(1-q^3)/(1-q)+a1