已知等差数列{an}中,公差d>0,其前n项和为Sn,且满足:a2*a3=45,a1+a4=14
问题描述：
已知等差数列{an}中,公差d>0,其前n项和为Sn,且满足:a2*a3=45,a1+a4=14
（1）求数列{an}通项公式.我会,答案是an=4n-3
（2）设数列bn=1/an*a(n+1)qi求数列{bn}前n项和Sn
第（2）题答案是sn=n/4n+1
答
1、略2、a(n+1)=4n+1bn=1/4*4/(4n-3)(4n+1)=1/4*[(4n+1)-(4n-3)]/[(4n-3)(4n+1)]=1/4*{(4n+1)/[(4n-3)(4n+1)]-(4n-3)/[(4n-3)(4n+1)]}=1/4*[1/(4n-3)-1/(4n+1)]所以Sn=1/4*[1/1-1/5+1/5-1/9+1/9-1/13+……+1/(4n-3)-...