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I know that

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

thanks to wolframalpha, but I don't understand the steps for simplification, and I can't seem to find any rules about factorial manipulations on google. Can someone explain this please?

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4  
Always use safety goggles. Always. Factorials blow up quite rapidly. –  Pedro Tamaroff Apr 2 '13 at 21:08
    
@Rokko: $(k+1)!-1+(k+1)(k+1)!\\=k(k+1)!+2(k+1)!-1\\=(k+2)(k+1)!-1\\=(k+2)!-1$ –  oldrinb Apr 2 '13 at 22:02
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4 Answers 4

up vote 4 down vote accepted

Ignore the $-1$, since they occur on both sides. Then you have:

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

Basically, it's just the distributive law.

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Now I understand. Thank you. –  Rokko Apr 2 '13 at 21:01
    
Except for (k+2)! = (k+2)(k+1)!. What does that have to do with ditributivity? –  Mythio Apr 3 '13 at 13:44
    
@Mythio $(a+b)c = ac+ bc$ is the distributive law. I'm just using $a=1$ and $b=k+1$ and $c=(k+1)!$. –  Thomas Andrews Apr 3 '13 at 14:16
    
@ThomasAndrews, I don't see how that helps in saying that (k+2)! = (k+2)(k+1)!. I can see how you used distributivity in the rest of the reasoning, but this first step seems to be some kind of rule involving the factorial operator? I'm confused how you can see that the first equality holds. –  Mythio Apr 3 '13 at 19:20
    
Ah, okay, that's just the recursive definition of factorial. $(n+1)!=(n+1)\cdot (n!)$. In this case, $n=k+1$. @Mythio –  Thomas Andrews Apr 3 '13 at 19:21
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Collect together the two terms that have a factor of $(k+1)!$:

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

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Assuming we are given only the task:

Simplify the following: $\quad (k+1)! - 1 + (k+1)(k+1)!$

Let $\color{blue}{\bf x = k+1}$. Then we have $$ \begin{align} x! - 1 + x(x!) & = x\;(x!) + 1\cdot x! -1 \\ \\ & = (x + 1)\;x! - 1 \quad\quad\quad\quad\quad\quad\tag{distributive property over addition} \\ \\ & = (\color{blue}{\bf x} + 1)! - 1 \\ \\ & = (\color{blue}{\bf k+1} + 1)! - 1 \\ \\ & = (k+2)! - 1 \end{align} $$

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Well, group like terms together. $(k+1)! + (k+1)(k+1)! = (k+2) (k+1)! = (k+2)!$.

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