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](/uploads/image/z/1846795-67-5.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E6%BB%A1%E8%B6%B32Sn%3Da%EF%BC%88n%2B1%EF%BC%89-2%5E%28n%2B1%29%2B1%2Cn%E2%88%88N%2C%E4%B8%94a1%2Ca2%2B5%2Ca3%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%971%2C%E6%B1%82a1%E7%9A%84%E5%80%BC2%2C%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
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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