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/(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.

1、1/2、1/4、1/7 1/4+1/7-0.25 25.6-2/7-4/7-1/7 计算(7+1)(7^2+1)(7^4+1)(7^8+1)(7^16+1) 6(7+1)(7^2+1)(7^4+1)(7^8+1)(7^16+1) -8×7/2×(-4/7)×1/4 1/4+1/4*1/7+1/7*1/10+...1/25*1/28=? 计算6(7+1)(7^+1)(7^4+1)(7^8+1)+1, 计算：6(7+1)(7^2+1)(7^4+1)(7^8+1)+1 1/7*2+1/7*4+1/7*8+1/7*16+1/7*32+1/7*64=?用拆项法解 1/1*4+1/4*7+1/7*10+.+1/2005*2008 1/1*4+1/4*7+1/7*10+.+1/2005*2008 1/1*4+1/4*7+1/7*10+.+1/52*55*+1/55*58=? 1/1×4+1/4×7+1/7×10+.+1/52×55+1/55×58 1/1*4+1/4*7+1/7*10+1/10*13+~+1/97*100 1如何计算1*1/4+4*1/7+7*1/10+...+2002*1/2005 计算1/1*4+1/4*7+1/7*10+...+1/2005*2008+1/2008*2011 1/1*4+1/4*7+1/7*10+1/10*13+…+1/298*301 1/1×4+1/4×7+1/7×10+...+1/22×25+1/25×28