So I have something like:
$\frac {k!}{(k-3)!3!}$
I'm going to add $\frac 12k(k-1)$ to this, and I want to obtain $\frac {(k+1)!}{(k-2)!3!}$ as the result. I'm having trouble with this since I need a $k+1$ in the numerator and a $k-2$ in the denominator. Any hints please? Thank you.
$\frac {k!}{(k-3)!3!}=\frac{k(k-1)(k-2)(k-3)}{6(k-3)}=\frac{k(k-1)(k-2)}{6}$ $\frac{k(k-1)(k-2)(k-3)}{6(k-3)}=\frac{k(k-1)(k-2)}{6}$ $\frac {k!}{(k-3)!3!}+\frac{1}{2}k(k-1)=\frac{k(k-1)(k-2)}{6}+\frac{1}{2}k(k-1)=\frac{k(k+1)(k-1)}{6}=\frac{(k+1)!}{3!}$.
On the other hand $\frac {(k+1)!}{(k-2)!3!}=\frac {(k+1)k(k-1)(k-2)!}{(k-2)!3!}=\frac{(k+1)!}{3!}$. So they are equal.