
(1)数列{an}的通项公式
∵an+1=2an+1,
∴an+1+1=2(an+1),
所以数列{an+1}是以a1+1=2为首项,以2为公比的等比数列,
∴an+1=2n,
即an=2n-1.
(2)数列{an}前n项和Sn
∵an=2n-1,
∴数列{an}前n项和Sn=a1+a2+a3+…+an
=(2-1)+(22-1)+(23-1)+…+(2n-1)
=(2+22+23+…+2n)-n
= 2(1-2n)1-2 -n
=2n+1-n-2.

(1)数列{an}的通项公式
∵an+1=2an+1,
∴an+1+1=2(an+1),
所以数列{an+1}是以a1+1=2为首项,以2为公比的等比数列,
∴an+1=2n,
即an=2n-1.
(2)数列{an}前n项和Sn
∵an=2n-1,
∴数列{an}前n项和Sn=a1+a2+a3+…+an
=(2-1)+(22-1)+(23-1)+…+(2n-1)
=(2+22+23+…+2n)-n
= 2(1-2n)1-2 -n
=2n+1-n-2.