1/(1*4)+1/(4*7)+1/(7*10)+1/(10*13)+1/(13*16)+1/(16*19)+1/(19*22)=?

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1/(1*4)+1/(4*7)+1/(7*10)+1/(10*13)+1/(13*16)+1/(16*19)+1/(19*22)=?1/(1*4)+1/(4*7)+1/(7*10)+1/(10*13)

1/(1*4)+1/(4*7)+1/(7*10)+1/(10*13)+1/(13*16)+1/(16*19)+1/(19*22)=?
1/(1*4)+1/(4*7)+1/(7*10)+1/(10*13)+1/(13*16)+1/(16*19)+1/(19*22)=?

1/(1*4)+1/(4*7)+1/(7*10)+1/(10*13)+1/(13*16)+1/(16*19)+1/(19*22)=?
此题初见可知,分母的表现形式为乘积且存在等差关系,分子为1,求几个数字之和
1.我们当然不应该硬算,题目有一定规律,从等差入手,把题目视为求数列之和,找出“通项”(每个乘号左边1,4,7...求出通项3n-2,每个乘号右边4,7,10.求出通项3n+1,)所以通项可视为1/(3n
-2)*(3n+1)
2求和,从以上通项可知1/(3n-2)*(3n+1)=[1/(3n-2)-1/(3n+1)]*1/3,然后就能把原式进行拆分,消去同项,进行简易求和,最后是(1-1/22)*1/3=7/22
楼主看在数学题难打的情况上给个采纳吧.