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

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

人气：245 ℃　时间：2020-02-03 08:30:08

解答

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)