(1999x2^1999-1998x2^1998)/(2000x2^2000)
=(1999/2-1998/4)/2000
=[(2000-1)/2-(2000-2)/4]/2000
=[1000-1/2-500+1/2]/2000
=1/4
(1999x2^1999-1998x2^1998)/2000x2^2000
=(1999x2-1998)/2000x4
=2000/8000
=1/4
这道题可以把1999看成a,则1998=a-1,2000=a+1
原始=[1999x2^1999-(1999-1)x2^(1999-1)]/(1999+1)x2^(1999+1)
=[1999x2^1999-(1999-1)/2x2^1999]/(1999+1)x2x2^1999
=[1999-(1999-1)/2]/(1999+1)x2
=[(1999+1)/2]/(1999+1)x2
=1/4