Siimplifying $(n+1)\cdot (n+1)! + (n+1)!-1$

Question

I'm trying to prove something and I have stumbled accross following line in my math book:

\begin{align*} & = (n+1)\cdot (n+1)! + (n+1)! - 1 \\ & = [(n+1) + 1]\cdot(n+1)! - 1 \\ & \end{align*}

But I'm unable to understand how that is possible. First, how can someone even see this relation in the first way? Second, why can we summarise (+1)⋅(+1)! in such a way?

Answer 1

All you're doing is factoring out $(n+1)!$ from the expression. It might be easier if we make a substitution. Let $p=(n+1)!$. Then your original expression is

$$(n+1)p + p - 1$$

Factor out $p$ throughout:

$$p\left( (n+1) + 1 \right) - 1$$

Simplify the inside:

$$p(n+2) - 1$$

Finally, bring back in the fact that $p=(n+1)!$:

$$(n+1)! \cdot (n+2) - 1$$

Answer 2

Note that $(n+2)!=(n+2)(n+1)!$ Your equation is just this with $1$ subtracted from each side and the $n+2$ written as $n+1+1$.

Answer 3

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