Question

数列{an}中,a1=1/6,an=(a1+a2+...+an-1)/(2+3+...+n),求1.a2,a3,a4;2.猜想数列{an}的通项公式并证明；

3.若Sn是数列{an}的前n项和，求极限lim(a1S2+a2S3+...+anSn+1)
速求回答，答题要规范的，若回答的好会继续追加分数，急急急!!!!!!!!!!3道小问都要回答的哟，步骤最好详细点!不胜感激.........

Answer 1

解 由题意可得an=(a1+a2+...+an-1)/(2+3+...+n),
a1=1/6=1/2*3
a2=(a1)/2=(1/6)/2 a2=1/12=1/3*4
a3=(a1+a2+)/(2+3)=(1/6+1/12)/5 a3=1/20=1/4*5
a4=(a1+a2+a3)/(2+3+4)=(1/6+1/12+1/20)/9=1/30=1/5*6
猜想an=1/(n+1)(n+2)
证明 a1=1/6=1/2*3
a2=1/12=1/3*4
a3=1/20=1/4*5
......
假设当n=k时 ak=1/(k+1)(k+2)成立
那么当n=k+1时
ak+1=(a1+a2+...+ak)/(2+3+。。。k+k+1)=(1/2*3+1/3*4+1/4*5+1/(k+1)(k+2))/(2+3+..k+1)
=(1/2-1/3+1/3-1/4-1/4+1/5+...+1/(k+1)-1/(k+2)）/（2+3+...+k+1)
=(1/2-1/k+2)/ (k+1+2)k/2
= 2*k/2(k+2)/k(k+2+1)
=1/(k+2)(k+2+1)
=1/(k+1+1)(k+1+2)
所以当n=k+1时也成立 n对一切自然数an=1/(n+1)(n+2) 均成立。
3）Sn=1/2*3+1/3*4+...+1/(n+1)(n+2)
=1/2-1/3+1/3-1/4+...+1/n+1-1/n+2
=1/2-1/n+2
=n/2n+4

追问

那第三问呢

追答

求：lim(a1S2+a2S3+...+anSn+1)
a1S2+a2S3+...+anSn+1
=1/2(1/3*4+1/4*5+....+1/(n+2)(n+3))
=1/2(1/3-1/4+1/4-1/5+...+1/n+2-1/n+3)
=1/2(1/3-1/n-3)
=1/6-1/3n-9
lim(a1S2+a2S3+...+anSn+1)
=lim(1/6-1/3n-9)
=1/6

Answer 2

an=(a1+a2+...+an-1)/(2+3+...+n)
则an*(2+n)(n-1)/2=a1+a2+...+a(n-1)
所以an*(n²+n-2)/2+an=a1+a2+...+an=Sn
即Sn=(n²+n)*an/2
1. a1=1/6
a2=S2-a1=(2²+2)*a2/2-1/6=3a2-1/6
a2=1/12
由S3=a1+a2+a3=1/4+a3=(3²+3)*a3/2=6a3
所以a3=1/20
由S4=a1+a2+a3+a4=1/6+1/12+1/20+a4=3/10+a4=(4²+4)*a4/2=10a4
所以a4=1/30
2. 猜想an=1/[(n+1)(n+2)]
由Sn=(n²+n)*an/2=[n(n+1)/2]*[Sn-S(n-1)]
得(n+2)Sn/n=(n+1)*S(n-1)/(n-1)=.....=3a1=3*1/6=1/2
所以Sn=n/[2(n+2)]=n(n+1)*an/2
an=1/[(n+1)(n+2)]
3. an*S(n+1)={1/[(n+1)(n+2)]}*(n+1)/[2(n+3)]
=1/[2(n+2)(n+3)]
=(1/2)[1/(n+2)-1/(n+3)]
所以a1S2+a2S3+...+anSn+1
=(1/2)[(1/3-1/4)+(1/4-1/5)+....+1/(n+2)-1/(n+3)]
=(1/2)[1/3-1/(n+3)]
=(1/6)*[n/(n+3)]
则lim(a1S2+a2S3+...+anSn+1)
=lim(1/6)*[n/(n+3)]
=1/6