在数列An中,A1=1,An+1=2An+2^n-1,求An。
1个回答
展开全部
a(n+1) = 2an +2^(n-1)
a(n+1)/2^(n+1) - an/2^n = 1/4
{an/2^n}是等差数列, d=1/4
an/2^n - a1/2^1= (n-1)/4
an/2^n = (2n+1)/4
an = (2n+1).2^(n-2)
a(n+1)/2^(n+1) - an/2^n = 1/4
{an/2^n}是等差数列, d=1/4
an/2^n - a1/2^1= (n-1)/4
an/2^n = (2n+1)/4
an = (2n+1).2^(n-2)
追问
是2^n,-1是单独的
追答
a(n+1) = 2an +2^n -1
a(n+1)/2^(n+1) - an/2^n = 1/2 - 1/2^(n+1)
an/2^n - a(n-1)/2^(n-1)= 1/2 - 1/2^n
an/2^n - a1/2^1 = (n-1)/2 - ( 1/2^2+1/2^3+...+1/2^n)
=(n-1)/2 - (1/2)[ 1- 1/2^(n-1) ]
an/2^n = (n-1)/2 + 1/2^(n-2)
an= (n-1).2^(n-1) + 1/2
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