等比数列 火速

来源:百度知道 编辑:UC知道 时间:2024/07/07 23:34:48
求数列[1/(1×3)]+[1/(3×5)]+[1/(5×7)]+……+[1/(2n-1)(2n+1)]的前n项和。

[1/(1×3)]+[1/(3×5)]+[1/(5×7)]+……+[1/(2n-1)(2n+1)]
=1/2(2/1*3+2/3*5.....+2/(2n-1)(2n+1)
=1-1/3+1/3-1/5+1/5-1/7......+1/(2n-1)-1/(2n+1)
=1-1/2n+1
=n/(2n+1)

Sn=(1-1/(2n+1))/2=n/(2n+1

2(1-1/3+1/3-1/5+1/5-1/7+…+1/(2n-3)-1/(2n-1))=2(1-1/(2n-1))

因为:1/(1×3)=(1-1/3)/2 1/(3×5)=(1/3-1/5)/2……
所以:Sn=(1-1/3)/2+(1/3-1/5)/2+…[1/(2n-1)-1/(2n+1)]/2
=[1-1/(2n+1)]/2
=[2n/(2n+1)]/2
=n/(2n+1)

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