ADD MATHS binomail

n! + (n+1)! + (n+2)!
= n! + (n+1)*n! + (n+2)(n+1)*n!
=[1+ (n+1) + (n+2)(n+1)] *n!
=[1 + n + 1 + n^2 + 3n + 2] *n!
=[n^2 + 3n +4] *n!


Is it correct???

2007-10-23 17:13:55 補充：
sorry, there is a mistake in the last step=[n^2 4n 4] * n!since (n^2 4n 4) can be factorize to become (n 2)^2therefore, the simpliest solution should be=[(n 2)^2 ]* n!

n!+(n+1)!+(n+2)!
= n! + n!(n+1) + n!(n+1)(n+2)
= n! (1 + n+1 + (n+1)(n+2))
= n! (n+2+n^2 + 3n +2)
= n!(n^2+4n+4)
= n!(n+2)^2