Simlifying [(k+1)! - 1] + (k+1)((k + 1)!)

Question

I'm afraid I've gotten a bit rusty on Math since I was last in university. I was looking at a problem in my text book that simplified

[(k+1)! - 1] + (k+1)((k + 1)!) down to (k+2)!-1 

I couldn't quite understand how this was done. Would anyone care to explain how this would work step by step? Thanks

Answer 1

Collect the two terms involving $(k+1)!$:

$$\begin{align*} \Big(\color{blue}{(k+1)!}\color{red}{-1}\Big)+(k+1)(k+1)!&=\color{blue}{(k+1)!}+(k+1)(k+1)!\color{red}{-1}\\ &=(k+1)!\Big(1+(k+1)\Big)-1\\ &=(k+1)!(k+2)-1\\ &=(k+2)!-1\;, \end{align*}$$

since

$$(k+2)!=(k+2)\Big((k+1)(k)\ldots(2)(1)\Big)=(k+2)(k+1)!\;.$$

Answer 2

$(k+1)! - 1 + (k+1)(k + 1)!$

$=(k+1)!-1+k(k+1)!+(k+1)!$

$=(k+1)!-1+(k+2)!+(k+1)!$

$=(k+2)!-1$

Answer 3

(k+1)! are common factors of the first and last terms