己知数列{an}满足a1=1,an+1=2n+1anan+2n (n∈N*),
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己知数列{an}满足a1=1,an+1=
2
(Ⅰ)∵数列{an}满足a1=1,an+1=
2n+1an an+2n (n∈N*), ∴ 2n+1 an+1= 2n an+1,即 2n+1 an+1− 2n an=1, ∴数列{ 2n an}是公差为1的等差数列. (Ⅱ)由(Ⅰ)可得 2n an= 2 a1+n−1=n+1, ∴an= 2n n+1. (Ⅲ)由(Ⅱ)知,bn=n(n+1)an=n•2n, ∴Sn=1×2+2×22+3×23+…+n•2n, 2Sn=22+2×23+…+(n-1)•2n+n•2n+1, 两式相减得:-Sn=2+22+…+2n-n•2n+1= 2(2n−1) 2−1-n•2n+1=(1-n)•2n+1-2, ∴Sn=(n-1)•2n+1+2.
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