Question

已知函数f(x)对任意实数p、q都满足:f(p+q)=f(p)×f(q)，且f(1)=3分之1，（1）当n属于N*时，求f(n)的表达式

已知函数f(x)对任意实数p、q都满足:f(p+q)=f(p)×f(q)，且f(1)=3分之1，（1）当n属于N*时，求f(n)的表达式，（2）设an=nf(n)(n属于N*)，sn是数列{an}的前n项和，求证:sn＜3／4

Answer 1

解：1) f(1)=1/3,
=> f(2)=f(1)×f(1)=(1/3)^2,
=> f(3)=f(1)×f(2)=(1/3)^3,
=> f(4)=f(1)×f(3)=(1/3)^4,
……
=> f(n)=f(1)×f(n-1)=(1/3)^n
2) an=n/3^n,
=> Sn=1/3+2/3^2+3/3^3+4/3^4+……+n/3^n, (1)
=> Sn/3= 1/3^2+2/3^3+3/3^4+……+(n-1)/3^n+n/3^(n+1), (2)
(1)-(2)得,
2Sn/3=1/3+1/3^2+1/3^3+……+1/3^n-n/3^(n+1)
<1/3+1/3^2+……+1/3^n=(1/3)×(1-1/3^n)/(1-1/3)=(1-1/3^n)/2,
=> Sn<(3/2)×(1-1/3^n)/2=3/4(1-1/3^n)<3/4,
=> Sn<3/4

Answer 2

(1)
f(p+q)=f(p)f(q)
put p=q=1/3
f(2)=f(1).f(1) =1/9
put p=2,q=1
f(3)=f(2)f(1) = (1/9)(1/3) =1/27
...
...
f(n) = f(n-1)f(1) =1/(3^n)
ie f(n) = 1/(3^n) 当n属于N*
(2)
an=nf(n)
Sn = 1(1/3)+2(1/9)+3(1/27)+...+n(1/3^n) (1)
(1/3)Sn = 1(1/9)+2(1/27)+.....+n(1/3^(n+1)) (2)
(1) -(2)
(2/3)Sn = 1(1/3)+1(1/9)+1(1/27)+...+1(1/3^n) - n(1/3^(n+1))
= (1/3)(1- (1/3)^n)/(1-1/3) -n(1/3^(n+1))
= (1- (1/3)^n)/2 - n(1/3^(n+1))
Sn = (3/4)(1- (1/3)^n)/2 - 3n(1/3^(n+1))/2
n->无穷
Sn ->3/4
=> Sn < 3/4