等差数列an=2n+3,求和：(1／a1a2)+(1／a2a3)+.+(1／anan+1)

原式=1/(5×7)+1/(7×9)+1/(9×11)+.+1/[(2n+3)(2n+5)]
=1/2[(1/5-1/7)+(1/7-1/9)+(1/9-1/11)+.+1/(2n+3)-1/(2n+5)]
=1/2[1/5-1/(2n+5)]
=n/(10n+25)

a1=2+3=5 d=2
1/anan+1=1/2(1/an-1/an-1)
原式=1/2[1/a1-1/a2+1/a2-1/a3+.......+1/an-1/an+1)=1/2(1/a1-1/an+1)=1/2[1/5-1/(2n+5)]

你是建一71届的吗？我也刚好再做这道题，哈
原式=1/(5×7)+1/(7×9)+1/(9×11)+....+1/anan+1
=1/2[(1/5-1/7+1/7-1/9+1/9-1/11+....+1/(2n+3)-1/(2n+5)]
=1/2[1/5-1/(2n+5)]
=n/(10n+25)