已知各项为整数的等差数列的前n项和为且a1a2a3a4=880则等差数列的通项公
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√Sn=√S1+(n-1)d√S2=√S1+d√S3=√S1+2d第2个式子两边平方a1+a2=a1+2(√a1)d+d^2第3个式子两边平方a1+a2+a3=a1+4(√a1)d+4d^2两个式子相
2S2=b2(a1+a2)=b1*q*(2a1+d)=32,b3S3=b3(a1+a2+a3)=b1*q²*(3a1+3d)=120,得d=2(都是正数),q=2.∴an=a1+d(n-1)
(1)、S2n-1=1/2an^2和an是各项均不为0的等差数列得S1=1/2a1^2=a1a1=2S3=1/2a2^2=3a2a2=6所以an=4n-2n为偶数时bn=1/2an-1=2n-3(2)
因为2Sn=an^2+n-4,所以2S(n-1)=a(n-1)²+n-1-4.两式相减2an=an^2-a(n-1)²+1,a(n-1)²=an^2-2an+1=(an-
S10=a1+a2+a3+……+a9+a10a1+a10=a2+a9=a3+a8=a4+a7=……所以S10=5(a4+a7)=100所以a4+a7=20a4+a7≥2根号a4*a7所以a4*a7≤(
AnA(n+1)在分子上?.是的话S1=A1=1=1/2(A1A2)=1/2(A2)A2=2S(n+1)=Sn+A(n+1)=1/2(AnA(n+1))+A(n+1)=1/2(A(n+1)A(n+2)
1.n=1时,2a1=2S1=a1²+1-4a1²-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=
sn=an(an+1)/2s(n-1)=a(n-1)(a(n-1)+1)/2两式相减an=an(an+1)/2-a(n-1)(a(n-1)+1)/2an^2-an-a^2(n-1)-a(n-1)=0(
n=1时,2a1=2S1=a1^2+1-4a1^2-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=an^2+n-4-a
此题是数列和均值不等式的结合S10=(a1+a10)/2*10=15a1+a10=3∴a3+a8=a1+a10=33[(1/a3)+(4/a8)]=(a3+a8)(1/a3+4/a8)=1+4+a8/
设等差数列{an}的公差为d(d≠0),则6a1+15d=60a1a21=a62,即6a1+15d=60a1(a1+20d) =(a1+5d) 2,解得:d=2a1=5,∴an=5
等差数列拿掉有限项后的公差不变,还是d拿掉m项后,原数列的第m+1项作为新数列的第一项.而原数列的第n+1项=a1+m*d(an=a1+(n-1)*d,这里n取m+1)所以首项am+1=a1+md
a1+a10=a2+a9=.a3+a8=10>=2更号下A3*A8A3*A8=25
∵{log2an}是公差为-1的等差数列∴log2an=log2a1-n+1∴an=2log2a1−n+1=a1•2−n+1∴S6=a1(1+12+…+132)=a1•1−1261−12=38,∴a1
根号Sn的通项公式是nSn=n^2an=Sn-Sn-1=n^2-(n-1)^2=2n-1
Sn、an、1成等差,则2an=Sn+1(n=1时,得a1=1),当n≥2时,有2a(n-1)=S(n-1)+1,则2an-2a(n-1)=an,即an/[a(n-1)]=2=常数,所以{an}是等比
由题意知2an=Sn+1/2,an>0,当n=1时,2a1=a1+1/2,解得a1=1/2,当n≥2时,Sn=2an-1/2,S(n-1)=2a(n-1)-1/2,两式相减得an=Sn-S(n-1)=
(1)由Sn,an,12成等差数列,可得2an=Sn+12,∴a1=12,a2=1(2)由2an=Sn+12可得,2Sn=4an-1(n≥1),∴2Sn-1=4an-1-1(n≥2)∴两式相减得2an
由题意2an=Sn+1/2Sn=2an-1/2n=1时,S1=a1a1=2a1-1/2a1=1/2S(n+1)-Sn=a(n+1)2a(n+1)-1/2-[2an-1/2]=a(n+1)a(n+1)=