Probability Question (1 Viewer)

Andy005 · Apr 14, 2017

Show that:

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

1729 · Apr 14, 2017

Andy005 said:
Show that:

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

$$\noindent $\begin{align*}(n+2)! + (n+1)! + n! &= (n+2)(n+1)n! + (n+1)n! + n! \\ &= n![(n+2)(n+1) + (n+1) + 1] \\ &= n!(n^2 + 4n +4) \\ &= n!(n+2)^2 \end{align*}$

Probability Question (1 Viewer)

Andy005

Member

1729

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)