已知数列{an}为等差数列,且a1=2,a1+a2+a3=12,通项公式为an=2n,令cn=1/an·an+1,求{c

Cn=1/[2n*(2n+2)]=1/[4n(n+1)]=1/4[1/n-1/(n+1)]
Sn=1/4[1-1/2+1/2-1/3+.+1/n-1/(n+1)]
=1/4[1-1/(n+1)]