1/1*2*3+1/2*3*4+...+98*99*100=
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解答：
1/(1*2*3)=(1/2)*[1/(1*2)-1/(2*3)]=(1/2)*[(1-1/2)-(1/2-1/3)]
1/(2*3*4)=(1/2)*[1/(2*3)-1/(3*4)]=(1/2)*[(1/2-1/3)-(1/3-1/4)]
.........
1/(98*99*100)=(1/2)*[1/(98*99)-1/(99*100)]=(1/2)*[(1/98-1/99)-(1/99-1/100)]
∴ 1/1*2*3+1/2*3*4+...+98*99*100
=(1/2)*[(1-1/2)-(1/2-1/3)]+(1/2)*[(1/2-1/3)-(1/3-1/4)]+........+(1/2)*[(1/98-1/99)-(1/99-1/100)]
=(1/2)*(1-1/2)] -(1/2)(1/99-1/100)]
=1/4-1/19800
=4949/19800
1/(1*2*3)=(1/2)*[1/(1*2)-1/(2*3)]=(1/2)*[(1-1/2)-(1/2-1/3)]
1/(2*3*4)=(1/2)*[1/(2*3)-1/(3*4)]=(1/2)*[(1/2-1/3)-(1/3-1/4)]
.........
1/(98*99*100)=(1/2)*[1/(98*99)-1/(99*100)]=(1/2)*[(1/98-1/99)-(1/99-1/100)]
∴ 1/1*2*3+1/2*3*4+...+98*99*100
=(1/2)*[(1-1/2)-(1/2-1/3)]+(1/2)*[(1/2-1/3)-(1/3-1/4)]+........+(1/2)*[(1/98-1/99)-(1/99-1/100)]
=(1/2)*(1-1/2)] -(1/2)(1/99-1/100)]
=1/4-1/19800
=4949/19800
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是1/(1*2*3)+1/(2*3*4)+...+1/(98*99*100)=吧
1/[n(n+1)(n+2)]={1/[n(n+1)]-1/[(n+1)(n+2)]}/2
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+……+1/(97*98*99)+1/(98*99*100)
=[1/(1*2)-1/(2*3)+1/(2*3)-1/(3*4)+.....+1/(97*98)-1/(98*99)+1/(98*99)
-1/(99*100)]/2
=[1/2-1/9900]/2
=4949/19800
1/[n(n+1)(n+2)]={1/[n(n+1)]-1/[(n+1)(n+2)]}/2
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+……+1/(97*98*99)+1/(98*99*100)
=[1/(1*2)-1/(2*3)+1/(2*3)-1/(3*4)+.....+1/(97*98)-1/(98*99)+1/(98*99)
-1/(99*100)]/2
=[1/2-1/9900]/2
=4949/19800
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