Question

在数列{an}中,a1=2,an+1=xan+x^n+1+(2-x)*2^n (n∈N*)x>0 1

在数列{an}中,a1=2,an+1=xan+x^n+1+(2-x)*2^n (n∈N*)x>0 1.求数列{an}的通项公式；
在数列{an}中,a1=2,a（n+1）=xan+x^（n+1）+(2-x)*2^n (n∈N*)x>0 1.求数列{an}的通项公式；

Answer 1

a(n+1)=xan+x^(n+1)+(2-x)*2^n
a(n+1)/x^(n+1) - an/x^n = 1+ (1/x)(2-x)(2/x)^n [ 两边除以x^(n+1)]
an/x^n - a(n-1)/x^(n-1) = 1+ (1/x)(2-x)(2/x)^(n-1)
an/x^n - a1/x^1= (n-1) + (1/x)(2-x) [ (2/x) + (2/x)^2 +...+ (2/x)^(n-1) ]
= (n-1) + (1/x)(2-x) (2/x) [ 1 - (2/x)^(n-1) ]/ [ 1-2/x]
=(n-1) - (2/x) [ 1 - (2/x)^(n-1) ]
an/x^n = (n-1) + (2/x)^n
an = (n-1).x^n + 2^n

追问

题目要求构造等差数列吖，求解

追答

an/x^n - a(n-1)/x^(n-1) = 1+ (1/x)(2-x)(2/x)^(n-1) (1)

a(n-1)/x^(n-1) - a(n-2)/x^(n-2) = 1+ (1/x)(2-x)(2/x)^(n-2) (2)
a(n-2)/x^(n-2) - a(n-3)/x^(n-3) = 1+ (1/x)(2-x)(2/x)^(n-3) (3)
....
....
a2/x^2 - a1/x^1 = 1+ (1/x)(2-x)(2/x)^1 (n-1)
(1)+(2)+(3)+...+(n-1)
an/x^n - a1/x^1= (n-1) + (1/x)(2-x) [ (2/x) + (2/x)^2 +...+ (2/x)^(n-1) ]
= (n-1) + (1/x)(2-x) (2/x) [ 1 - (2/x)^(n-1) ]/ [ 1-2/x]
=(n-1) - (2/x) [ 1 - (2/x)^(n-1) ]
an/x^n = (n-1) + (2/x)^n
an = (n-1).x^n + 2^n