simplifiying an expression $(n + 1)! − 1 + (n + 1) \cdot (n + 1)!$

Question

I've been stuck on this one problem and I have a problem on the process simplifying this equation so that it is $(n + 2)! − 1.$

$$(n + 1)! − 1 + (n + 1) \cdot (n + 1)!$$

If anyone could shed some light on this matter, please..

Basically how would you get from $(n + 1)! − 1 + (n + 1) \cdot (n + 1)!$ to $(1 + n + 1) \cdot(n + 1)! − 1$

i understand thanks – Anonymous Platypus Apr 13 '15 at 18:11 — Anonymous Platypus, Apr 13 '15 at 18:11
its says to wait 5 mins – Anonymous Platypus Apr 13 '15 at 18:15 — Anonymous Platypus, Apr 13 '15 at 18:15

score 2 · Answer 1 · answered Apr 13 '15 at 18:07

2

$$(n+1)!+(n+1)\cdot (n+1)!=(n+1)![1+(n+1)]=\cdots$$

answered Apr 13 '15 at 18:07

lab bhattacharjee

274,582

score 1 · Accepted Answer · answered Apr 13 '15 at 18:08

It comes from factoring out the quantity $(n+1)!$. If we let $x = (n+1)!$ then we'd have $$x-1+(n+1)x$$ and factoring out $x$ yields $$x(1+(n+1))-1$$ Hence, $$\begin{align}(n + 1)! − 1 + (n + 1)\cdot(n + 1)! = (n+1)![1+(n+1)]-1 \\ = (n+1)![n+2]-1 \\ = (n+2)!-1\end{align}$$

simplifiying an expression $(n + 1)! − 1 + (n + 1) \cdot (n + 1)!$

2 Answers2