1/4+1/28+1/70+.+1/9700=?
问题描述：
1/4+1/28+1/70+.+1/9700=?
答
1/4+1/28+1/70+.........+1/9700=1/3[(1-1/4)+(1/4-1/7)+(1/7-1/10)+.........+(1/97-1/100)]=1/3[1-1/100]=0.33
答
人家都答了,
答
传说中的裂项消元法...
答
1/4=1/3(1/1-1/4)
1/28=1/3(1/4-1/7)
1/70=1/3(1/7-1/10)
..........
1/9700=1/3(1/97-1/100)
1/4+1/8+1/70+...+1/9700
=1/3(1/1-1/4+1/4-1/7+1/7-1/10......+1/97-1/100)
=1/3(1-1/100)
=33/100
答
1/4+1/28+1/70+.+1/9700
=1/(1*4)+1/(4*7)+1/(7*10)+...+1/(97*100)
=(1/3)[(1/1)-(1/4)]+(1/3)[(1/4)-(1/7)]+(1/3)[(1/7)-(1/10)]+...+(1/3)[(1/97)-(1/100)]
=(1/3)[1-(1/4)+(1/4)-(1/7)+(1/7)-(1/10)+...+(1/97)-(1/100)]
=(1/3)[1-(1/100)]
=(1/3)*(99/100)
=33/100.