等差数列｛1\an｝满足a1=1,公差d=2,求a1a2+a2a3+……+anan+1的和

1/a1=1
d=2
所以1/an=(2n-1)
所以原式=1/1*3+1/3*5+……+1/(2n-1)(2n+1)
=(1/2)(1-1/3)+(1/2)(1/3-1/5)+……+(1/2)[1/(2n-1)-1/(2n+1)]
中间正负抵消
=(1/2)[1-1/(2n+1)]
=n/(2n+1)