设数列{an}的前n项和为Sn满足2Sn=an+1-2n+1+1,n∈N*,且a1,a2+5,a3成等差数列.
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设数列{an}的前n项和为Sn满足2Sn=an+1-2n+1+1,n∈N*,且a1,a2+5,a3成等差数列.
(1)求a1的值;
(2)求数列{an}的通项公式.
(1)求a1的值;
(2)求数列{an}的通项公式.
(1)在2Sn=an+1-2n+l+1中,
令n=1得:2S1=a2-22+1,即a2=2a1+3 ①
令n=2得:2S2=a3-23+1,即a3=6a1+13 ②
又2(a2+5)=a1+a3 ③
联立①②③得:a1=1;
(2)由2Sn=an+1-2n+l+1,得:
2Sn+1=an+2-2n+2+1,
两式作差得an+2=3an+1+2n+1,
又a1=1,a2=5满足a2=3a1+21,
∴an+1=3an+2n对n∈N*成立.
∴an+1+2n+1=3(an+2n).
∴an+2n=3n.
则an=3n-2n.
令n=1得:2S1=a2-22+1,即a2=2a1+3 ①
令n=2得:2S2=a3-23+1,即a3=6a1+13 ②
又2(a2+5)=a1+a3 ③
联立①②③得:a1=1;
(2)由2Sn=an+1-2n+l+1,得:
2Sn+1=an+2-2n+2+1,
两式作差得an+2=3an+1+2n+1,
又a1=1,a2=5满足a2=3a1+21,
∴an+1=3an+2n对n∈N*成立.
∴an+1+2n+1=3(an+2n).
∴an+2n=3n.
则an=3n-2n.
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