1∵an和bn公共项组成新数列cn∴ an=bn得n=1或n=1/2(舍去)∴只有n=1时an和bn有公共项∴a1=b1=cn=2∴sn=2n2 (1)∵a1=1，an+1 =(n+2)sn/n∴ sn=[n/(n+2)]a(n+1)∴s(n-1)=[(n-1)/(n+1)]an相减得：an==[n/(n+2)]a(n+1)-[(n-1)/(n+1)]an∴[1/(n+2)]a(n+1)=2[1/(n+2)/(n+1)]an∴sn/n=[1/(n+2)]a(n+1); s(n-1)/(n-1)=[1/(n+2)/(n+1)]an∴sn/n=2s(n-1)/(n-1)∴数列sn/n为以2为公比的等比数列 (2)由上知：s(n+1)/(n+1)=s1*2^*n=2^n∴s(n+1)=(n+1)2^n∵s(n-1)=[(n-1)/(n+1)]an=(n-1)*2^(n-2)∴an=(n+1)*2^(n-2)∴sn=4an

1. An={2,5,8,11,…,3n-1}Bn={2,4,8,…,2^n},3m-1=2^n,3m=(2^n+1), ∴2^+1是3的倍数, ∴ Cn={2,8,32,128,…,2^(2n-1)}, Cn=2^(2n-1)=4^n/2 数列{Cn}是首项为2,公比为4的等比数列∴ Sn=2(4^n-1)/32. (1) nA(n+1)=(n+2)Sn, n[S(n+1)-Sn])=(n+2)Sn,nS(n+1)=2(n+1)Sn,∴ [S(n+1)/(n+1)]/(Sn/n)=2, ∴ 数列{Sn/n}为等比数列(2) ∵ 数列{Sn/n}是首项为S1=1,公比为2的等比数列∴ S(n+1)/(n+1)=1·2^n, S(n+1)=(n+1)·2^n…①又S(n+1)/(n+1)=2(Sn/n)=2A(n+1)/(n+2)=2^n,∴ A(n+1)=(n+2)·2^(n-1),从而An=(n+1)·2^(n-2),∴ 4An=(n+1)·2^n…②,由①,②得S(n+1)=4An