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