【答案】
分析:(1)先計算f(0)<0,f(1)>0,且f(x)在R上的圖象是一條連續(xù)曲線,根據(jù)零點存在定理得f(x)在(0,1)內(nèi)有零點,再根據(jù)其導(dǎo)數(shù)為正,得出f(x)在(0,1)上是增函數(shù),f(x)在(0,1)內(nèi)只有一個零點,而a
n是函數(shù)f(x)=x
3+n
2x-1(n∈N
+)的零點,從而證明出0<a
n<1;
(2)分兩部分進行證明.先證明左邊的不等式,由(1)知0<a
n<1,得a
n>

,利用放縮法及裂項法可得a
1+a
2+…+a
n>1-

+

-

+

-

+…+

=

;再證明右邊的不等式,由于a
n=

,當n≥2時,可得a
1+a
2+…+a
n<

+

+

-

+

-

+…+

-

=1+

-

<

.綜上可得

.
解答:解:(1)∵f(0)=-1<0,f(1)=n
2>0,且f(x)在R上的圖象是一條連續(xù)曲線,
∴f(x)在(0,1)內(nèi)有零點,
∵f′(x)=3x
2+n
2>0,∴f(x)在(0,1)上是增函數(shù),f(x)在(0,1)內(nèi)只有一個零點,
而a
n是函數(shù)f(x)=x
3+n
2x-1(n∈N
+)的零點,
∴0<a
n<1;
(2)先證明左邊的不等式,因a
n3+n
2a
n-1=0,由(1)知0<a
n<1,
∴a

<a
n,即1-n
2a
n=a

<a
n.
∴a
n>

,∴a
1+a
2+…+a
n>

+

+…+

①
∵a
n>

≥

=

,
∴a
1+a
2+…+a
n>1-

+

-

+

-

+…+

=

,
再證明右邊的不等式,由于f(

)=

+

-1=-

<0,f(

)=

>0,
∴

<a
1<

,
由(1)知,0<a
n<1,且a
n3+n
2a
n-1=0,
∴a
n=

,
∵當n≥2時,a
1+a
2+…+a
n<

+

+

-

+

-

+…+

-

=1+

-

<

,
∴當n∈N
*時,a
1+a
2+…+a
n<

,
綜上,

.
點評:本小題主要考查零點、函數(shù)單調(diào)性的應(yīng)用、數(shù)列與函數(shù)的綜合、不等式的證明等基礎(chǔ)知識,考查運算求解能力,考查化歸與轉(zhuǎn)化思想.屬于較難題.