(1)由已知推得f
k(x)=(n-k+1)x
n-k,從而有f
k(1)=n-k+1
(2)證法1:當(dāng)-1≤x≤1 時(shí),F(xiàn)(x)=x
2n+nc
n1x
2(n-1)+(n-1)c
n2x
2(n-2)+…+(n-k+1)c
nkx
2(n-k)+…+2c
nn-1x
2+1
當(dāng)x>0時(shí),F(xiàn)′(x)>0
所以F(x)在[0,1]上為增函數(shù)
因函數(shù)F(x)為偶函數(shù),所以F(x)在[-1,0]上為減函數(shù)
所以對(duì)任意的x
1,x
2∈[-1,1],|F(x
1)-F(x
2)|≤F(1)-F(0)
F(1)-F(0)=C
n0+nc
n1+(n-1)c
n2+…+(n-k+1)c
nk+…+2c
nn-1=nc
nn-1+(n-1)c
nn-2+…+(n-k+1)c
nn-k+…+2c
n1+c
n0∵(n-k+1)c
nn-k=(n-k)c
nn-k+c
nk=nc
n-1k+c
nk(k=1,2,3,…,n-1)
F(!)-F(0)=n(c
n-11+c
n-12+..+c
n-1k-1)+(c
n1+c
n2+…+c
nn-1)+c
n0=n(2
n-1-1)+2
n-1=2
n-1(n+2)-n-1
因此結(jié)論成立.
證法2:當(dāng)-1≤x≤1 時(shí),F(xiàn)(x)=x
2n+nc
n1x
2(n-1)+(n-1)c
n2x
2(n-2)+…+(n-k+1)c
nkx
2(n-k)+…+2c
nn-1x
2+1
當(dāng)x>0時(shí),F(xiàn)′(x)>0
所以 F(x)在[0,1]上為增函數(shù)
因函數(shù) F(x)為偶函數(shù)
所以 F(x)在[-1,0]上為減函數(shù)
所以對(duì)任意的x
1,x
2∈[-1,1],|F(x
1)-F(x
2)|≤F(!)-F(0)
F(!)-F(0)=c
n0+nc
n1+(n-1)c
n2+…+(n-k+1)c
nk+…+2c
nn-1又因F(1)-F(0)=2c
n1+3c
n2+…+kc
nk-1+…+nc
nn-1+c
n0所以2[F(1)-F(0)]=(n+2)[c
n1+c
n2+…+c
nk-1+…+c
nn-1]+2c
n0F(1)-F(0)=
[c
n1+c
n2+…+c
nk-1+…+c
nn-1]+c
n0=
(2n-2) +1=2n-1(n+2)-n-1因此結(jié)論成立.
證法3:當(dāng)-1≤x≤1時(shí),F(xiàn)(x)=x
2n+nc
n1x
2(n-1)+(n-1)c
n2x
2(n-2)+…+(n-k+1)c
nkx
2(n-k)+…+2c
nn-1x
2+1
當(dāng)x>0時(shí),F(xiàn)′(x)>0
所以F(x)在[0,1]上為增函數(shù)
因函數(shù)F(x)為偶函數(shù)
所以F(x)在[-1,0]上為減函數(shù)
所以對(duì)任意的x
1,x
2∈[-1,1],|F(x
1)-F(x
2)|≤F(!)-F(0)
F(!)-F(0)=c
n0+nc
n1+(n-1)c
n2+…+(n-k+1)c
nk+…+2c
nn-1由x[(1+x)
n-x
n]=x[c
n1x
n-1+c
n2x
n-2+…+c
nkx
n-k+…+c
nn-1+1]=c
n1x
n+c
n2x
n-1+…+c
nkx
n-k+1+…+c
nn-1x
2+x
對(duì)上式兩邊求導(dǎo)得(1+x)
n-x
n+nx(1+x)
n-1-nx
n=nc
n1x
n-1+(n-1)c
n2x
n-2+…+(n-k+1)c
nkx
n-k+…+2c
nn-1x+1
F(x)=(1+x
2)
n+nx
2(1+x
2)
n-1-nx
2n∴F(1)-F(0)=2
n+n2
n-1-n-1=(n+2)2
n-1-n-1.
因此結(jié)論成立.