已知等差数列前几项和Sn=3n的平方 2,求通项an

S9/T9=9a5/9b5=a5/b5=63/12=21/4S8/T8=4(a4+a5)/[4(b4+b5)]=(a4+a5)/(b4+b5)=56/11S7/T7=7a4/7b4=a4/b4=49/

http://zhidao.baidu.com/question/88231937.html?fr=qrl&cid=983&index=2S1=a1=-(2/3),S2+1/S2+2=a2,因为S2=

a1=1+根号2S3=9+3根号2a2=3+根号2an=2n-1+根号2sn=n^+n根号2bn=sn/n=n+根号假设bn,bk,bm成等比则m+n=2k,mn=k^2解得m=n,不符题意

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

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

S1=a1=-(2/3),S2+1/S2+2=a2,因为S2=(a1+a2),所以S2+1/S2+2=S2-a1=S2+2/3,解得S2=-(3/4),同理,S3+1/S3+2=a3=S3-S2=S3

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-

Sn=n(A1+An)/2Tn=n(B1+Bn)/2Sn/Tn=(A1+An)/(B1+Bn)然后n代2n-1A2n-1+A1=2AnBn同理S2n-1/T2n-1=An/Bn=7(2n-1)/(2n

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

由等差数列的性质Sn=na1+n(n-1)d/2=dn2/2+(a1-d/2)n=An2+Bn即A=d/2B=a1-d/2同样地Tn=nb1+n(n-1)p/2=pn2/2+(b1-p/2)n=Cn2

{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

an=S1n=1an=Sn-Sn-1n>1得到a1=8an=5(n)^2+3n-5(n-1)^2-3(n-1)=10n-5+3=10n-2n>1因为恰好n=1时也满足上面的公式所以an=10n-2n>

Sn=-2n^2-nS(n-1)=-2(n-1)^2-(n-1)an=Sn-S(n-1)=-2n^2-n+2(n-1)^2+(n-1)=2[(n-1)^2-n^2]-1=-4n+1a1=-3an是以-

解：①当n=1时a1=S1=2②当n≥2时an=Sn-Sn-1Sn=3n^2-nSn-1=3（n-1)²-(n-1)所以an=6n-4=2+6（n-1）带入n=1得到a1=2符合①综上所述a

S9=(a1+a9)/2*9=(2a5)/2*9=9a5同理T9=9b5a5／b5=S9:T9=21/4

当n≥2时,可以化为Sn-S(n-1)=-2Sn×S(n-1),两边同除以Sn×S(n-1),得1／Sn-1／S(n-1)=2所以｛1/Sn｝是以2为首项,2为公差的等差数列即1/Sn=2nSn=1/

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

证::n=1,a1=s1=4n>1an=Sn-Sn-1Sn=n^2+3nSn-1=(n-1)^2+3(n-1)an=2n+2经验证n=1满足通项n>1an-an-1=2,由等差数列定义可知,数列{an

n≥2时,a(n)=S(n)-S(n-1)=(2n²+3n)-[2(n-1)²+3(n-1)]=4n+1当n=1时,a1=S1=2×1+3×1=5,也适合上面式子∴a(n)=4n+

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