Question

已知数列{an}满足a1=1，an+1=2an+n-1(Ⅰ)求证:数列{an+n...

已知数列{an}满足a1=1，an+1=2an+n-1 (Ⅰ)求证:数列{an+n}是等比数列; (Ⅱ)求数列{an}的通项和前n项和Sn.

Answer 1

解答:解:(Ⅰ)由数列{an}满足a1=1，an+1=2an+n-1，变形为an+1+(n+1)=2(an+n).
∴数列{an+n}是等比数列，其中首项为a1+1=2，公比为2;
(II)由(I)可得:an+n=2×2n-1，∴an=2n-n.
∴Sn=
2(2n-1)
2-1
-
n(n+1)
2
=2n+1-2-
n(n+1)
2
.