等差数列an=2n-1,求数列{1/an*a n+1}的前n项和Sn

1/an*a(n+1)=1/(2n-1)(2n+1)=(1/2)[1/(2n-1)-1/(2n+1)]
所以Sn=1/1*3+1/3*5+1/5*7+.+1/(2n-1)(2n+1)
=(1/2)(1-1/3+1/3-1/5+.+1/(2n-1)-1/(2n+1))
=(1/2)(1-1/(2n+1))
=n/(2n+1)

1/[an.a(n+1)]
=1/[(2n-1)(2n+1)]
=(1/2)[1/(2n-1) -1/(2n+1)]
Sn = (1/2)[ 1- 1/(2n+1)]
= n/(2n+1)

{an}是1,3,5,7,9......
数列{1/an*an+1}，表示1/(1*3),1/(3*5),1/(5*7),1/(7*9)......
所以1/(an*an+1)=1/2*(1/an - 1/an+1)
所以Sn=1/2*(1/a1-1/a2+1/a2-1/a3+1/a3-1/a4......+1/an-1/an+1)
=1/2*(1/a1-1/an+1)
=1/2*{1-1/[2*(n+1)-1]}
=1/2*[1-1/(2n+1)]=n/(2n+1)