Question

已知函数f(x)=，数列{a_n}满足a₁=1，a_n+1=f(a_n)(n∈N^*)．

(1)求证：数列{}是等差数列；

(2)记S_n(x)=,求S_n(x)．

Answer 1

解：（1）由已知得：a_n+1=

∴

∴{}是首项为1，公差d=3的等差数列 

（2）由（1）得=1+(n-1)3=3n-2

∴S_n(x)=x+4x²+7x³+…+(3n-5)x^n-1+(3n-2)xⁿ

当x=1,S_n(1)=1+4+7+…+(3n-2)

= 

当x≠1,0时，S_n(x)=x+4x²+7x³+…+(3n-5)x^n-1+(3n-2)xⁿ

xS_n(x)=x²+4x³+7x⁴+…+(3n-5)xⁿ+(3n-2)xⁿ⁺¹

(1-x)S_n(x)=x+(3x²+3x³+…+3xⁿ)-(3n-2)xⁿ⁺¹

=x+-(3n-2)xⁿ⁺¹

∴S_n(x)=

=

=

=

当x=0时，S_n(0)=0也适合.

综上所述，x=1,S_n(1)=

x≠1,S_n(x)=.