48*（1/（9-4）+1/（16-4）+1/（25-4）+...+1/(10000-4))

来源：百度知道 编辑：UC知道 时间：2024/05/10 11:34:47

解：原式=48[1/(9-4)+1/(16-4)+1/(25-4)+...+1/(10000-4)]
=48[1/(3²-2²)+1/(4²-2²)+1/(5²-2²)+....+1/(100²-2²)]
=48[1/(3-2)(3+2)+1/(4-2)(4+2)+1/(5-2)(5+2)+....+1/(100-2)(100+2)]
=48{1/4[1/(3-2)-1/(3+2)]+1/4[1/(4-2)-1/(4+2)]+1/4[1/(5-2)-1/(5+2)]+....+1/4[1/(100-2)-1/(100+2)]}
=12[(1+1/2+1/3+....+1/98)-(1/5+1/6+1/7+....+1/102)]
=12[(1+1/2+1/3+1/4)-(1/99+1/100+1/101+1/102)]
=12[(1-1/101)+(1/2-1/102)+(1/3-1/99)+(1/4-1/100)]
=12(100/101+25/51+32/99+6/25)
=1200/101+100/17+128/33+72/25
=30500/(17*101)+5576/(25*33)
=34736492/1416525