问题补充:
已知数列{an}的前n项和为Sn,满足Sn+2n=2an
(1)证明:数列{an+2}是等比数列.并求数列{an}的通项公式an;
(2)若数列{bn}满足bn=log2(an+2),设Tn是数列的前n项和.求证:.
答案:
证明:(1)由Sn+2n=2an得?Sn=2an-2n
当n∈N*时,Sn=2an-2n,①
当n=1?时,S1=2a1-2,则a1=2,
则当n≥2,n∈N*时,Sn-1=2an-1-2(n-1).②
①-②,得an=2an-2an-1-2,
即an=2an-1+2,
∴an+2=2(an-1+2)
∴,
∴{an+2}是以a1+2为首项,以2为公比的等比数列.
∴an+2=4?2n-1,
∴an=2n+1-2.
(2)证明:由bn=log2(an+2)==n+1,
得,
则,③
??④
③-④,得
=
=
=,
所以?.
解析分析:(1)由Sn+2n=2an,知Sn=2an-2n.当n=1?时,S1=2a1-2,则a1=2,当n≥2时,Sn-1=2an-1-2(n-1),故an=2an-1+2,由此能够证明数列{an+2}是等比数列.并能求出数列{an}的通项公式an.(2)由bn=log2(an+2)==n+1,得,故,由此利用错位相减法能够求出Tn,并证明.
点评:本题考查等比数列的证明和求数列{an}的通项公式an,.解题时要认真审题,注意构造法和错位相减法的合理运用.
已知数列{an}的前n项和为Sn 满足Sn+2n=2an(1)证明:数列{an+2}是等比数列.并求数列{an}的通项公式an;(2)若数列{bn}满足bn=log2