已知函数y=f(x)是定义在R上恒不为0的单调函数,对任意的x,y∈R,总有f(x)f(y)=f(x+y)成立,若数列{an}的n项和为Sn,且满足a1=f(0),f(an+1)=1/f(3n+1−2an)(n∈N*),则Sn=_.

问题描述:

已知函数y=f(x)是定义在R上恒不为0的单调函数,对任意的x,y∈R,总有f(x)f(y)=f(x+y)成立,若数列{an}的n项和为Sn,且满足a1=f(0),f(an+1)=

1
f(3n+1−2an)
(n∈N*),则Sn=______.

因为任意的x,y∈R,总有f(x)f(y)=f(x+y)成立,
所以f(0)f(0)=f(0),即f(0)•(f(0)-1)=0,
解得f(0)=1,即a1=1,
又f(an+1)•f(3n+1-2an)=1,即f(an+1+3n+1-2an)=f(0),
所以an+1+3n+1-2an=0,
则an+1+3n+1+2×3n+1=2an+2×3n+1,,即

an+1+3n+2
an+3n+1
=2,
所以数列{an+3n+1}是首项为10,公比为2的等比数列,
则an+3n+1=10×2n-1,即an=5×2n-3n+1
所以Sn=5×
2(1−2n)
1−2
-
32(1−3n)
1−3
=2n+1
3n+2+11
2

故答案为2n+1
3n+2+11
2