设数列An的前n项和为Sn,已知A1=A2=1,Bn=nSn+(n+2)An,数列Bn是公差为d的...

n2Sn?1＝n(n?1)，∴n+1nSn?nn?1Sn?1＝1，对n≥2成立． …（3分）又1+11S1=1，∴{n+1nSn}是首项为1，公差为1的等差数列．n+1nSn＝1+(n?1)?1…（5分）∴Sn＝n2n+1…（6分）（2）证明：bn＝Snn3+3n＝1(n+1)(n+3)＝12(1n+1?1n+3)…（...

则nSn+（n+2）an=4n 则Sn=4-(n+2)/nan=4-an-2an/n 则S(n+1)=4-a(n+1)-2a(n+1)/(n+1)两式相减 得a(n+1)=an-a(n+1)+2an/n-2a(n+1)/(n+1)即[2+2/(n+1)]a(n+1)=(1+2/n)an 即(2n+4)/(n+1)a(n+1)=((n+2)/n)an 即a(n+1)=(n+1)/2na...

设数列｛an｝的前n项和为Sn,已知a1=1，且a（n+1）=（n+2）/n sn ,求证：数列{sn/n}是等比数列。证明：因为A(n+1) =(n+2)/n * Sn 所以Sn =n*A(n+1) / (n+2)S(n-1) = (n-1)*An / (n+1)所以An = Sn -S(n-1) = n/(n+2) *A(n+1) - (n-1)/(n+1...

2×…×a2a1＝nn?1×n?1n?2×…×21＝n，n≥2，∴an=4n（n≥2）．∵a1=4满足上式，∴an=4n．∵{an}是等差数列，Sn＝(4+4n)n2＝2n(n+1)，∴2n（n+1）（bn+2bn-bn+12）+bn+1bn=0，bn+2bn?bn+12bn+1bn＝?12n(n+1)，∴...

证：由a1=1，an+1=[(n+2)/n]Sn（n=1,2,3)知a2=3a1 S2/2=4a1/2=2 S1/1=1 ∴(S2/2)/(S1/1)=2 又an+1=Sn+1-Sn（n=1，2，3，…）则Sn+1-Sn=[(n+2)/n]Sn ∴nSn+1=2（n+1）Sn (Sn+1/n+1)/(Sn/n)=2（n=1，2，3，…）故数列{Sn/n}是首项为1，公比...

1/(sn-1)-1/sn=4 所以1/sn为公差为-4的等差数列 1/sn=1/s1+(n-1)*(-4) 因为1/s1=1/a1=1 1/sn=5-4n sn=1/(5-4n)an=sn-sn-1=1/(5-4n)-1/(5-4(n-1))an=[(9-4n)-(5-4n)]/[(5-4n)(9-4n)]an=4/[(5-4n)(9-4n)] (n=>2)

--n-1/1-n n+1/n Sn --n/n-1 Sn-1=1 所以数列{（（n+1）/n）Sn是等差数列 S1=a1. n+1/n Sn=2S1+n-1=n 所以Sn=n^2/n+1 证明：bn=1/n(n+1)=1/n -1/n+1 b1+b2+b3+...+bn=1-1/2+1/2-1/3+...1/n-1/n+1 =n/n+1 <1 ...

(1)an+1=(n+2)/nSn,即S(n+1)-Sn=(n+2)/nSn,化简可得S(n+1)/(n+1)=2(Sn/n),即证得数列{Sn/n}是等比数列；(2)由（1）可知Sn=n*2^(n-1),可求出an=(n+1)*2^(n-2),即可证得S(n+1)=4an.

（1）n≥2时，an=Sn-S(n-1)=n²+n-(n-1)²-(n-1)=2n.又a1=2，所以an=2n.（2）1/Sn=1/[n(n+1)]=1/n-1/(n+1).所以Tn=(1-1/2)+(1/2-1/3)+...+[1/n-1/(n+1)]=1-1/(n+1)<1.

nS(n+1)=(n+2)Sn S(n+1)/Sn = (n+2)/n Sn/S(n-1) = (n+1)/(n-1)Sn/S1 = (n+1)n/2 Sn = (n+1)n/2 an = Sn -S(n-1)= n