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等差数列an=2n+3,求和:(1/a1a2)+(1/a2a3)+.+(1/anan+1)
人气:397 ℃ 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)
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