(1)由an=(3n+Sn)可得Sn=2an-3n,故an+1=Sn+1-Sn=2an+3 ∵a1=(3+S1),∴a1=3,∴a2=9,a3=21; (2)证明:由待定系数法得an+1+3=2(an+3) 又a1+3=6≠0 ∴数列{an+3}是以6为首项,2为公比的等比数列. ∴an+3=6×2n-1, ∴an=3(2n-1). (3)由(2)可得bn=n2n-n, ∴Bn=1×21+2×22+3×23+…+n×2n-(1+2+3+…+n) ① ∴2Bn=1×22+2×23+3×24+…+n×2n+1-2(1+2+3+…+n) ② ①-②得,-Bn=2+(22+23+…+2n)+ 化简可得Bn=2+(n-1)2n+1- 假设数列{an}存在构成等差数列的四项依次为:am、an、ap、aq(m<n<p<q) 则3(2m-1)+3(2q-1)=3(2n-1)+3(2p-1)∴2m+2q=2n+2p. 上式两边同除以2m,则1+2q-m=2n-m+2p-m ∵m、n、p、q∈N*,且m<n<p<q, ∴上式左边是奇数,右边是偶数,相矛盾. ∴数列{an}不存在构成等差数列的四项. |