设数列an的前n项和为sn,满足2Sn=a（n+1）-2^(n+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
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设数列an的前n项和为sn,满足2Sn=a（n+1）-2^(n+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
设数列an的前n项和为sn,满足2Sn=a（n+1）-2^(n+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
设数列an的前n项和为sn,满足2Sn=a（n+1）-2^(n+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）在2Sn=a-2^(n+1)+1中,
令n=1得：2S1=a2-2²+1,
令n=2得：2S2=a3-2³+1,
解得：a2=2a1+3,a3=6a1+13
又2（a2+5）=a1+a3
解得a1=1
（2）由2Sn=a-2^(n+1)+1,
2S=a-2^(n+2)+1得a=3a+2^(n+1),
又a1=1,a2=5 也满足a2=3a1+2,
所以a=3an+2n对n∈N*成立
∴an+1+2^(n+1)=3（an+2^n）,又a1=1,a1+2=3,
∴an+2^n=3^n,
∴an=3^n-2^n；
2Sn=a(n+1)-2^(n+1)+1 (1)
n=1
2a1= a2-2^2+1
a2 = 2a1+3
n=2
2(a1+a2) = a3 -2^3 +1
2(a1+2a1+3)= a3-7
6a1+6=a3-7
a3= 6a1+13
a1，a2+5，a3成等差数列
=>
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2Sn=a(n+1)-2^(n+1)+1 (1)
n=1
2a1= a2-2^2+1
a2 = 2a1+3
n=2
2(a1+a2) = a3 -2^3 +1
2(a1+2a1+3)= a3-7
6a1+6=a3-7
a3= 6a1+13
a1，a2+5，a3成等差数列
=>
a1+a3 = 2(a2+5)
a1+(6a1+13) = 2(2a1+3+5)
7a1+13=4a1+16
a1=1
2S(n-1)=an-2^(n)+1 (2)
(1)-(2)
2an = a(n+1) -an -2^n
a(n+1) = 3an+2^n
a(n+1) + 2^(n+1) = 3(an + 2^n)
[a(n+1) + 2^(n+1)]/(an + 2^n)=3
(an + 2^n)/(a1 + 2^1)=3^(n-1)
an + 2^n = 3^n
an = 3^n -2^n
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