2x5/1+5x8/1+8x11/1+.2006x2009/1+2009x2012/1 用裂项法计算
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2x5/1+5x8/1+8x11/1+.2006x2009/1+2009x2012/1 用裂项法计算
由于[1/n]-[1/(n+3)]= 3/[n(n+3)]
所以有1/[n(n+3)]=(1/3)[(1/n)-1/(n+3)]
所以原式1/2*5+1/5*8+1/8*11+...+1/2006*2009+1/2009*2012
=(1/3)[(1/2-1/5)+(1/5-1/8)+(1/8-1/11)+...+(1/2006-1/2009)+(1/2009-1/2012)]
=(1/3)[1/2-1/2012]
=(1/3)*(1005/2012)
=335/2012
所以有1/[n(n+3)]=(1/3)[(1/n)-1/(n+3)]
所以原式1/2*5+1/5*8+1/8*11+...+1/2006*2009+1/2009*2012
=(1/3)[(1/2-1/5)+(1/5-1/8)+(1/8-1/11)+...+(1/2006-1/2009)+(1/2009-1/2012)]
=(1/3)[1/2-1/2012]
=(1/3)*(1005/2012)
=335/2012
2x5/1+5x8/1+8x11/1+.2006x2009/1+2009x2012/1 用裂项法计算
裂项法1/2X5+1/5x8+1/8x11.1/2009x2012=?
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