Question

已知数列An中，A1=1,An+1*An-1=An*An-1+An的平方（n∈N*，n≥2），且An+1/An=kn+1.

(1)求证:k=1(2)设g(x)=An*x的n-1次方/(n-1)！,f(x)是数列g(x)的前n项和,求f(x)的解析式

Answer 1

(1)
转换得an+1/an=(an/an-1)+1=(an-1/an-2)+1+1=(a2/a1)+n-1
又a2/a1=k+1
->(an+1/an)=n+k=(k*n)+1
->k=1
(2)
(an/an-1)=n-1+1=n
->an=n*an-1=n*(n-1)*an-2=...=n!
g(x)=An*x^(n-1)/(n-1)!=n*x^(n-1)
f(x)=n+n*2^(n-1)+n*3^(n-1)+...+n*n^(n-1)
=n(1+2^(n-1)+...+n^(n-1))