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1/(12)+1/(23)+1/(34)+1/(45)的简便计算的过程

来源：百度知道编辑：UC知道时间：2024/05/30 20:21:03

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原式=1-1/2+1/2-1/3……
即1/(n*(n+1))=1/n-1/(n+1)
所以原式=1-1/5=4/5.

1/1*2+1/2*3+1/3*4+1/4*5
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)
=1-1/2+1/2+1/3-1/3+1/4-1/4-1/5
=1-1/5
=4/5

1/1*2=1-1/2
1/2*3=1/2-1/3
1/3*4=1/3-1/4
1/4*5=1/4-1/5
原式=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)
=1-1/2+1/2+1/3-1/3+1/4-1/4-1/5
=1-1/5
=4/5

1-1/2+1/2-1/3+1/3-1/4+1/4-1/5=1-1/5=4/5

(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1) 1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+100) 1+1/(1+2)+1/(1+2+3)+-------+1/(1+2+3+----+100) 1+1/1+2+1/1+2+3+...+1/1+2+3...+2000 1+1/1+2+1/1+2+3.........+1/1+2+3.....100 1*(1/1+2)*(1/1+2+3)*~~~*(1/1+2+~~~2005)=? (1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000) (1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8) (1-1/2^2)*(1-1/3^2)*(1-1/4^2).......(1-1/100^2) 1+1/2+1+1/3+1+1/4+......+1/100=?