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；

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那第三问呢求：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-9lim(a1S2+a2S3+...+anSn+1)=lim(1/6-1/3n-9)=1/6