Question

1/1*2*3+1/2*3*4+...+1/8*9*10+1/9*10*11的答案是多少？

Answer 1

1/[n*(n+1)(n-2)]
=1/[n*（n+1）]-1/[n*(n+2)]
=1/n-1/(n+1)-1/2(1/n-1/(n+2))

1/1*2*3+1/2*3*4+...+1/8*9*10+1/9*10*11
=1-1/2-1/2(1-1/3)+1/2-1/3-1/2(1/2-1/4)......1/9-1/10-1/2(1/9-1/11)
=[1-1/2+1/2-1/3+....+1/9-1/10]-1/2[1-1/3+1/2-1/4+1/3-1/5....+1/9-1/11]
=9/10-1/2[1+1/2-1/10-1/11]
=9/10-1/2*144/110
=9/10-72/110
=27/110

Answer 2

1/2*3*4=1/4(1/1*2*3*4-1/2*3*4*5)
原式=1/1*2*3+1/4(1/1*2*3*4-1/2*3*4*5+1/2*3*4*5- . . . - 1/9*10*11*12)
=1/1*2*3+1/4(1/1*2*3*4- 1/9*10*11*12)
=1/6+1/48* 247/495
=4207/32760

Answer 3

27/110

追问

可是我看到别人的答案是21/11，能给过程吗？

追答

1/(n-1)n(n+1)= 2n/2(n-1)nn(n+1)=1/2*n*(n+1-n+1)/(n-1)nn(n+1)=0.5*(1/(n-1)n-1/n(n+1))