分析:(1)解法一:通過函數(shù)的導(dǎo)數(shù),判斷函數(shù)的單調(diào)性,求出最大值即可求a
1,a
2的值;
解法二:利用函數(shù)的導(dǎo)數(shù),求出函數(shù)的最值,推出a
1,a
2的值.
(2)利用(1)解法求出n≥3時(shí)函數(shù)的最大值,即可求數(shù)列{a
n}的通項(xiàng)公式;
(3)利用分析法以及二項(xiàng)式定理直接證明:對(duì)任意n∈N
*(n≥2),都有
an≤成立.
解答:解:(1)解法1:∵
fn′(x)=nxn-1(1-x)2-2xn(1-x)=xn-1(1-x)[n(1-x)-2x]-------(1分)
當(dāng)n=1時(shí),f
1'(x)=(1-x)(1-3x)
當(dāng)
x∈[,1]時(shí),f
1'(x)≤0,即函數(shù)f
1(x)在
[,1]上單調(diào)遞減,
∴
a1=f1()=,--------------------------------------------------(3分)
當(dāng)n=2時(shí),f
2'(x)=2x(1-x)(1-2x)
當(dāng)
x∈[,1]時(shí),f
2'(x)≤0,即函數(shù)f
2(x)在
[,1]上單調(diào)遞減,
∴
a2=f2()=---------------------------------------------------(5分)
【解法2:當(dāng)n=1時(shí),
f1(x)=x(1-x)2,則
f1′(x)=(1-x)2-2x(1-x)=(1-x)(1-3x)當(dāng)
x∈[,1]時(shí),f
1'(x)≤0,即函數(shù)f
1(x)在
[,1]上單調(diào)遞減,∴
a1=f1()=,
當(dāng)n=2時(shí),
f2(x)=x2(1-x)2,則
f2′(x)=2x(1-x)2-2x2(1-x)=2x(1-x)(1-2x)
當(dāng)
x∈[,1]時(shí),f
2'(x)≤0,即函數(shù)f
2(x)在
[,1]上單調(diào)遞減,∴
a2=f2()=】
(2)令f
n'(x)=0得x=1或
x=,
∵當(dāng)n≥3時(shí),
∈[,1]且當(dāng)
x∈[,)時(shí)f
n'(x)>0,
當(dāng)
x∈(,1]時(shí)f
n'(x)<0,-----------------(7分)
故f
n(x)在
x=處取得最大值,
即當(dāng)n≥3時(shí),
an=fn()=()n()2=
,-------(9分)
當(dāng)n=2時(shí)(*)仍然成立,
綜上得
an=-------------------------------------(10分)
(3)當(dāng)n≥2時(shí),要證
≤,只需證明
(1+)n≥4,-------------------(11分)
∵
(1+)n=+()+…+()n≥1+2+•≥1+2+1=4∴對(duì)任意n∈N
*(n≥2),都有
an≤成立.-----------------(14分)
點(diǎn)評(píng):本題考查數(shù)列與函數(shù)的導(dǎo)數(shù)的應(yīng)用,考查分析問題解決問題的能力,數(shù)列通項(xiàng)公式的求法,二項(xiàng)式定理的應(yīng)用,考查計(jì)算能力轉(zhuǎn)化思想的應(yīng)用.