1/(1*4)+1/(4*7)+1/(7*10)+……+1/(97*100)

问题描述:

1/(1*4)+1/(4*7)+1/(7*10)+……+1/(97*100)

很简单,你注意观察,1/(1*4),相当于三分之一倍的(1-1/4),后面的同理,所以原式可以变为1/3(1-1/100),立面的东西都可以相互抵消掉,最终等于33/100.

1/(1*4)+1/(4*7)+1/(7*10)+……+1/(97*100)
=1-3/4+3/4-5/7+5/7-7/10+…-65/97+67/100
=1-67/100
=33/100
其中,97为1~97,差值为3的等差,所以间隔(97-1)/3,同样的差值为2的就是
2*(97-1)/3=65

1/(1*4)+1/(4*7)+1/(7*10)+……+1/(97*100)
=3*[(1/1-1/4)+(1/4-1/7)+(1/7-1/10)+……+(1/97-1/100)]
=3*[1-1/100]
=3*[99/100]

1/3*(1-1/4)+1/3*(1/4-1/7)+...+1/3(1/97-1/100)=1/3(1-1/100)=3*99/100

1/(1*4)+1/(4*7)+1/(7*10)+……+1/(97*100)
=1/3(1-1/4+1/4-1/7+.+1/97-1/100)
=1/3(1-1/100)
=33/100

1/(1*4)+1/(4*7)+1/(7*10)+……+1/(97*100)
因为1/(1*4)=(1/3)*(1/1-1/4)
1/(4*7)=(1/3)*(1/4-1/7)
1/(7*10)=(1/3)*(1/7-1/10)
……
1/(97*100)=(1/3)*(1/97-1/100)
所以原式=(1/3)(1/1-1/4+1/4-1/7+1/7-1/10+……+1/97-1/100)
=(1/3)(1/1-1/100)
=33/100

1/3(1-1/4+1/4-1/7+1/7-1/10+.....+1/97-1/100)=1/3(1-1/100)=33/100对吗