Question

一道初中数学题！具体看补充！急！好的加分！

试通过计算说明/K(K+)=(/K)-/(K+)，然后计算1/X(X+1)+[1/(X+1)(X+2)]+...+[1/(X+99)(X+100)].好的加分！

Answer 1

1/X(X+1)+[1/(X+1)(X+2)]+...+[1/(X+99)(X+100)]=1/X-1/(X+1)+1/(X+1)-1/(X+2)]+...+1/(X+99)-1/(X+100)=1/X-1/(X+100)=100/X(X+100)

Answer 2

1/X(X+1)+[1/(X+1)(X+2)]+...+[1/(X+99)(X+100)]
=[1/X-1/(X+1)]+[1/(X+1)-1/(X+2)]+...+[1/(X+99)-1/(X+100)].
=1/X-1/(X+100)
=100/[X(X+100)]

从以上计算显然可以看出
1/K(K+1)=(1/K)-1/(K+1)

Answer 3

1/2*3=1/2-1/3,1/3*4=1/3-1/4.......
1/X(X+1)+[1/(X+1)(X+2)]+...+[1/(X+99)(X+100)]=1/X-1/X+1+1/X+1-1/X+2+........-1/X+100=1/X-1/X+100

Answer 4

1/k(k+1)
=[(k+1)-k]/k(k+1)
=(k+1)/k(k+1)-k/k(k+1)
=1/k-1/(k+1)

所以原式=1/x-1/(x+1)+1/(x+1)-1/(x+2)+……+1/(x+99)-1/(x+100)
=1/x-1/(x+100)

=100/(x²+100x)