已知数列{an}满足an=1/(2n-7)(2n-5),其前n项和为Sn 求证求数列{5/（10Sn+1）}
问题描述：
已知数列{an}满足an=1/(2n-7)(2n-5),其前n项和为Sn 求证求数列{5/（10Sn+1）}
答
an=1/(2n-7)(2n-5),
=(1/2)*[1/(2n-7)-1/(2n-5)]
Sn=a1+a2+a3+...+an
=(1/2)*[-1/5+1/3-1/3+1-1-1+1+...+1/(2n-7)-1/(2n-5)]
=(1/2)*[-1/5-1/(2n-5)]
10Sn=5*[-1/5-1/(2n-5)]
=-1-1/(2n-5)
10Sn+1=-1/(2n-5)
5/（10Sn+1）=-5*(2n-5)
设5/（10Sn+1）为bn
则b(n+1)=-5*（2n-3)
b(n+1)-bn=-5*（2n-3)+5*（2n-5）
=-10
所以数列{5/（10Sn+1）}为等差数列