已知f(x)=a0+a1x+a2x^2+…+anx^n(x∈N*),记f1(x)=f'(x),f2(x)=f1'(x),…,fn(x)=f'n-1(x),且f(0)=f1(0)=f2(0)=…=fn(0)=1.(Ⅰ)求f(x)(Ⅱ)求证f(1)≤3-1/2^n-1

Question

已知f(x)=a0+a1x+a2x^2+…+anx^n(x∈N*),记f1(x)=f'(x),f2(x)=f1'(x),…,fn(x)=f'n-1(x),且f(0)=f1(0)=f2(0)=…=fn(0)=1.
(Ⅰ)求f(x)
(Ⅱ)求证f(1)≤3-1/2^n-1

Answer 1

1.
f(0)=a0=1
f1(0)=a1=1
f2(0)=2a2=1
..
fk(0)=k!ak=1
知ak=1/k!,01*2*2*..*2=2^(n-1)
所以
f(1)=1+1/1!+1/2!+...+1/n!
≤1+1+1/(2^1)+1/(2^2)+..+1/(2^(n-1))
=1+2(1-(1/2)^n)
=3-1/(2^(n-1))
等号成立当且仅当n