nbn

c(n)=a(n)b(n)=(2n-1)3^n,s(n) = (2*1-1)3 + (2*2-1)3^2 + (2*3-1)3^3 + ... + [2*(n-1)-1]3^(n-1) + [2n-1]3^n,3s(n) = (2*1-1)3^2 + (2*2-1)3^3+...+[2(n-1)-1]3^n + [2n-1]3^(n+1),2s(n) = 3s(n) - s(n) = -(2*1-...

（1）令n=1，得a1=S1=2a1-1，解得a1=1，当n≥2时，an=Sn-Sn-1=2（an-an-1），整理，得an=2an-1，∴an＝2n?1．∵数列{bn}满足b1=1，nbn+1=（n+1）bn，∴bn+1n+1＝bnn，∴{bnn}是首项为1的常数列，∴bnn＝1，∴bn=n．（2）∵数列{bn}的前n项和为Qn，∴Qn＝1+2+3+…+n＝n(n+1)2，∵Tn=Sn+Qn，∴Tn=2?2n?1?1+n(n?1)2=2n?1+n(n+1)2，当n=1时，λT1≥T2，得λ≥3，当n=2时，λT2≥T3，得λ≥136，猜想：当λ≥3时，3Tn≥Tn+1．证明：3Tn?Tn+1＝3[2n?1+n(n+1)2]-[2n+1?1+(n+1)(n+2)2]=2n+n-3≥0．综上所述，λ存在最小值3，使不等式λTn≥Tn+1成立．

(n+1)[b(n+1)]^2-nbn^2+b(n+1).bn=0[(n+1)b(n+1)-nbn].[b(n+1) + bn]=0case 1:b(n+1) + bn=0bn/b(n-1) = -1bn = (-1)^(n-1)case 2:(n+1)b(n+1)-nbn =0b(n+1)/bn= n/(n+1)bn/b(n-1) = (n-1)/nbn/b1 = 1/nbn =1/n