已知等差数列{an} 若Sn=a S2n=b 求S3n

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

设公差为d.Sn=Smna1+n(n-1)d/2=ma1+m(m-1)d/2(m-n)a1+(m²-n²-m+n)d/2=0(m-n)a1+[(m+n)(m-n)-(m-n)]d/

S(9)=9a(5)那么a(5)=2,sn=(a1+an)*n/2=(a5+a(n-4))*n/2=240n=15第二题先假设n是个奇数,An/Bn=a((n+1)/2)/b((n+1)/2)=(7n

数列｛Sn/n｝构成一个公差为2的等差数列,∴Sn/n=2n,∴Sn=2n^2,∴a3=S3-S2=18-8=10.

加一个条件,an是正项数列2Sn=(an)²+n-42S(n-1)=[a(n-1)]²+(n-1)-4n>=2则2an=2Sn-2S(n-1)=(an)²-[a(n-1)

解an是等差数列∴2a4=a3+a5∵a3+a4+a5=12∴a4+2a4=12∴a4=4∵s7=(a1+a7)×7÷2=2a4×7÷2=4×728再答：利用2a(n+1)=an+a(n+2)再问：a

Sn-S(n-m)=A(n-m+1)+A(n-m+2)+……+A(n-m+m)=b共m项A(n-m+1)=A1+(n-m)dA(n-m+2)=A2+(n-m)d……A(n-m+m)=An=Am+(n-

条件不充分,根据这一条只能求出a9=20,（a1和d两个未知数）没法对任意的n求Sn...再问：求S17

条件不足,无解,但注意高级魔法师的前n相和的求法只适用于等比数列公比小于1的情况,和此题无关

sn=an^2+bns(n-1)=a(n-1)^2+b(n-1)两式作差,由：sn-s(n-1)=an可证．

{an}是等差数列,a2=a1+da3=a1+2d....an=a1+(n-1)da(2n-1)=a1+(2n-2)da1+a(2n-1)=2a1+(2n-2)d2an=2a1+2(n-1)d=2a1

{an}和{bn}公差分别设为d1、d2Sn=na1+n(n-1)d1/2sn'=nb1+n(n-1)d2/2Sn/sn'=[2a1+(n-1)d1]/[2b1+(n-1)d2]=(2n+3)/(3n

因为是等差的,所以和的个数是偶数的话,和=中间两项相加*个数/2也就是说=（a4+a5）*8/2=72（8就是一共有8个数相加,a4、a5为中间两项）如果和的个数是奇数的话,和=中间一项*2*（个数+

S1/a1=1S2/a2-S1/a1=(2+d)/(1+d)-1=d/(1+d)S3/a3-S1/a1==(3+3d)/(1+2d)-1=(2+d)/(1+2d)2*d/(1+d)=(2+d)/(1+

∵等差数列{a[n]},S[n]=[(a[n]+1)/2]^2∴4S[n]=a[n]^2+2a[n]+1∵4S[n+1]=a[n+1]^2+2a[n+1]+1∴将上面两式相减,得：4a[n+1]=a[

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∴

等差数列,An=A1+(n-1)d=A1+7n-7=18A1+7n=25,A1=25-7nSn=n(A1+An)/2=n(A1+18)/2=20A1n+18n=40(25-7n)n+18n=407n^

由S(n+1)/S(n)=(4n+2)/(n+1),可得a(n+1)/S(n)=S(n+1)/S(n)-1=(3n+1)/(n+1),所以S(n)=(n+1)/(3n+1)*a(n+1)以n-1代替n

当n≥2时，an=Sn-Sn-1=2n+1∴a2=5，a3=7∴d=7-5=2a1=1+2+a=3+a∵{an}为等差数列∴a1=a2-d=3=3+a∴a=0故答案为：0

等差数列数列的性质a1+a[2n-1]=2an因为S[2n-1]=[(2n-1)(a1+a[2n-1])]/2=(2n-1)anT[2n-1]=[(2n-1)(b1+b[2n-1])]/2=(2n-1