1/11*12*13+1/12*13*14+1/13*14*15等于多少?
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∵1/[(x-1)x(x+1)]
=[(x+1)-(x-1)]/[2(x-1)x(x+1)]
=1/[2x(x-1)]-1/[2x(x+1)]
=[x-(x-1)]/[2x(x-1)]-[(x+1)-x]/[2x(x+1)]
=(1/2)[1/(x-1)-1/x]-(1/2)[1/x-1/(x+1)]
=(1/2)[1/(x-1)-2/x+1/(x+1)],
∴依次令上述等式中的x为12、13、14,可依次得:
1/(11×12×13)=(1/2)(1/11-2/12+1/13),
1/(12×13×14)=(1/2)(1/12-2/13+1/14),
1/(13×14×15)=(1/2)(1/13-2/14+1/15),
将上述三个等式左右分别相加,得:
1/(11×12×13)+1/(12×13×14)+1/(13×14×15)
=(1/2)(1/11-1/12-1/14+1/15)
=(1/2)[(12-11)/(11×12)-(15-14)/(14×15)]
=(1/2)[1/(11×12)-1/(14×15)]
=(1/12)[1/(11×2)-1/(7×5)]
=(1/12)(1/22-1/35)
=(1/12)(35-22)/(22×35)
=(1/12)×[13/(11×70)]
=13/(12×770)
=13/9240
=[(x+1)-(x-1)]/[2(x-1)x(x+1)]
=1/[2x(x-1)]-1/[2x(x+1)]
=[x-(x-1)]/[2x(x-1)]-[(x+1)-x]/[2x(x+1)]
=(1/2)[1/(x-1)-1/x]-(1/2)[1/x-1/(x+1)]
=(1/2)[1/(x-1)-2/x+1/(x+1)],
∴依次令上述等式中的x为12、13、14,可依次得:
1/(11×12×13)=(1/2)(1/11-2/12+1/13),
1/(12×13×14)=(1/2)(1/12-2/13+1/14),
1/(13×14×15)=(1/2)(1/13-2/14+1/15),
将上述三个等式左右分别相加,得:
1/(11×12×13)+1/(12×13×14)+1/(13×14×15)
=(1/2)(1/11-1/12-1/14+1/15)
=(1/2)[(12-11)/(11×12)-(15-14)/(14×15)]
=(1/2)[1/(11×12)-1/(14×15)]
=(1/12)[1/(11×2)-1/(7×5)]
=(1/12)(1/22-1/35)
=(1/12)(35-22)/(22×35)
=(1/12)×[13/(11×70)]
=13/(12×770)
=13/9240
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