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

所以n/1*2*3+(n-1)/2*3*4+…1/n*(n+1)(n+2)
=（n+1)/[2*1*(1+1)]-（n+1)/[2(n+1)*(n+2)]-1/(1+1)+1/(n+2）
=(n-1)/4+1/[2(n+2)]