题目

数列｛an｝及fn（x)=a1x+a2x^2+…+anx^n,fn(-1)=n•(-1)^n,n=1,2,3…
1,求a1,a2,a3的值 2,求数列通项 3,证,1/3小于等于fn(1/3)小于1

提问时间：2020-10-21

答案

fn(-1) = a1(-1)^1 + a2(-1)^2+.+an(-1)^n = n(-1)^n
fn-1(-1) = a1(-1)^1 + a2(-1)^2+.+an-1(-1)^(n-1)=(n-1)(-1)^(n-1)
上减下得an(-1)^n=(2n-1)(-1)^n
an=2n-1
带入1,2,3,a1=1,a2=3,a3=5
fn(1/3)=1*1/3 + 3*(1/3)^2 + 5*(1/3)^3 +.+ 2n-1*(1/3)^n
1/3*fn(1/3)=1*(1/3)^2 + 3*(1/3)^3 +.+(2n-3)*(1/3)^n + 2n-1*(1/3)^n+1
上下错位相减得2/3*fn(1/3)=1*(1/3) + 2*(1/3)^2 + 2*(1/3)^3 +.+2*(1/3)^n - 2n-1*(1/3)^n+1
=1/3 - 2n-1*(1/3)^n+1 + 2[(1/3)^2 + (1/3)^3 +.+(1/3)^n]
=1/3 - 2n-1*(1/3)^n+1 + 1/3*[1- (1/3)^n-1]
fn(1/3) =1 - (n+1)*(1/3)^n 显而易见小于1
并且由fn(1/3)递增可知fn(1/3)>f1(1/3) fn(1/3)>1/3