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/...