2/1*2*3+2/2*3*4+……+2/11*12*13
=2（1/1*2*3+1/2*3*4+……+1/11*12*13）
到這里先停一下.
根據(jù)1/x*(x+1)*(x+2)
=(x+2-x-1)/x*(x+1)*(x+2)
=1/x*(x+1)-1/x*(x+2)
因此原式=2(1/1*2 -1/1*3+1/2*3-1/2*4+……+1/11*12-1/11*13)
=2(1/1*2 +1/2*3+……+1/11*12-1/1*3-1/2*4-……-1/11*13)
到這里再停一下.
根據(jù)1/x*(x+1)
=(x+1-x)/x*(x+1)
=1/x-1/(x+1)
以及1/x*(x+2)
=0.5*[x+2-2/x*(x+2)]
=0.5*[1/x-1/(x+2)]
因此原式=2[1-1/2+1/2-1/3+……+1/11-1/12-0.5(1-1/3+1/2-1/4+……+1/11-1/13)]
=2[1-1/12-0.5(1+1/2-1/12-1/13)]
=2[11/12-0.5*209/156]
=2[286/312-209/312]
=77/156