己知数列{an}满足a1=1,an+1=2n+1anan+2n (n∈N*), (Ⅰ)证明数列{ 2nan }是等差数列; (Ⅱ)求数列{an)的通项公式; (Ⅲ)设bn=n(n+1)an 求数列{bn}的前n项和Sn.
己知数列{an}满足a1=1,an+1=
(n∈N*),
(Ⅰ)证明数列{
}是等差数列;
(Ⅱ)求数列{an)的通项公式;
(Ⅲ)设bn=n(n+1)an 求数列{bn}的前n项和Sn.
2n+1an |
an+2n |
(Ⅰ)证明数列{
2n |
an |
(Ⅱ)求数列{an)的通项公式;
(Ⅲ)设bn=n(n+1)an 求数列{bn}的前n项和Sn.
数学人气:996 ℃时间:2019-11-10 16:58:02
优质解答
(Ⅰ)∵数列{an}满足a1=1,an+1=2n+1anan+2n (n∈N*),∴2n+1an+1=2nan+1,即2n+1an+1−2nan=1,∴数列{2nan}是公差为1的等差数列.(Ⅱ)由(Ⅰ)可得2nan=2a1+n−1=n+1,∴an=2nn+1.(Ⅲ)由(Ⅱ)知,bn=...
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