求数列1/2x5,1/5x8,1/8x11...,1/(3n-1)(3n+2)...的前n项和

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求数列1/2x5,1/5x8,1/8x11...,1/(3n-1)(3n+2)...的前n项和
=1/3[1/(3n-1)-1/(3n+2)]
=1/3(1/2-1/5)+1/3(1/5-1/8)+1/3……1/3[1/(3n-1)-1/(3n+2)]
=1/3[1/2-1/5+1/5-1/8+1/8-……+1/(3n-1)-1/(3n+2)]
.
=n/[2(3n+2)]
=1/3[1/(3n-1)-1/(3n+2)]这步是如何得到的?

这叫裂项求和法基本裂项式是

如有不懂请追问望采纳

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