Question

已知数列{an}及fn(x)=a1x+a2x^2+……+anx^n，fn(-1)=(-1)^n*n，n=1，2，3……(1)求a1，a2，a3，的值 (2...

已知数列{an}及fn(x)=a1x+a2x^2+……+anx^n，fn(-1)=(-1)^n*n，n=1，2，3……(1)求a1，a2，a3，的值 (2)求数列{an}的通项公式(3)求证：fn(1/3)<1

Answer 1

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

数列{an}的通项公式为an=2n-1

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fn(1/3)于1(因为n为1，2，3……）
所以fn(1/3)<1

Answer 2

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赞同14| 评论