因为,(n-1)n(n+1)(n+2)+1=[(n-1)(n+2)][n(n+1)]+1=(n^2+n-2)(n^2+n)+1
=[(n^2+n-1)-1][(n^2+n-1)+1]+1=(n^2+n-1)^2
所以,根号下{[(n-1)n(n+1)(n+2)+1]/4}=(n^2+n-1)^2
所以,原式=(1999^2+1999-1)/2=(2000*1999-1)/2=3997999/2