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
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；
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；
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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
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
1年前 追问
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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
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