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Exercise: Prove that (n + 1)! - n! = n(n!) for any n $\ge$ 1

Given Answer:

I will skip the basic step since I understand that part.

(n + 2)! - (n + 1)! = (n + 1)!(n + 2) - n!(n + 1) I understand this line
But, I don't understand starting at this next line
= n(n + 1)! + 2(n + 1)! - n(n!) - n!
= n(n + 1)! + (n + 1)! - n(n!) + (n + 1)! - n!
= n(n + 1)! + (n + 1)! - n(n!) + n(n!)
= n(n + 1)! + (n + 1)!
= (n + 1)!(n + 1)

If anyone can help me to understand this then that would be great!

Thanks,

Tony

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Is any body aware of any good tutorials that show examples of solving factorials by induction? –  Tony Oct 21 '13 at 21:02

1 Answer 1

up vote 1 down vote accepted

$(n+2)! = (n+2)(n+1)! = n(n+1)! + 2(n+1)!$, by the distributive property.


The inductive step would be to assume that $(n + 1)! - n! = n(n!) $ holds for some $n > 1$.

We then want to show that $\color{blue}{(n+2)!} -\color{red}{(n+1)!} = (n+1)(n+1)!$

We begin by rearranging...

$$\color{blue}{(n+2)!} -\color{red}{(n+1)!} =\color{blue}{(n+2)}(n+1)! - \color{red}{(n+1)}(n!)$$ $$=\color{blue}{n}(n+1)! + \color{green}{2}(n+1)! - \color{red}{n}(n!) -\color{red}{1}(n!) $$

Notice that I changed the blue $2$ into green. That is so we can have $$=\color{black}{n}(n+1)! + \color{green}{1}(n+1)! - \color{black}{n}(n!) + \color{green}{1}(n+1)! -\color{black}{1}(n!) $$

The two terms on the right are $(n+1)! - n!$, which equals $\color{red}{n(n!)}$ by hypothesis. So now we have... $$=\color{black}{n}(n+1)! + \color{black}{1}(n+1)! - \color{black}{n}(n!) + \color{red}{n(n!)} $$

The two terms on the right cancel. $$=\color{blue}{n}(n+1)! + \color{blue}{1}(n+1)!$$

Using the distributive property (factoring out a $(n+1)!$, if you will) ... $$\color{blue}{(n+1)}(n+1)!$$

So the claim holds for $n+1$.

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Okay, thanks. I see that now but, what happened to the 2 in the next line and, is n! = (n + 1)! - n!? Hmm, is 2(n + 1)! equal to both of the (n + 1)! in the next line? –  Tony Oct 21 '13 at 20:54
    
The $2(n+1)!$ splits into two terms of $(n+1)!$ in the next line. You would do well to write all this out and use colored pencils to keep track of the various terms/factors. –  The Chaz 2.0 Oct 21 '13 at 21:12
    
Thanks again. I have copied your answer and made some notes for it also. –  Tony Oct 21 '13 at 21:48
    
Hi again. Am I correct in assuming that (n + 1)! - n! is equal to n(n!). If this is correct do you mind showing how this is worked out? I am sorry for all of the questions but I am really struggling with this exercise. The study guide I have does not really show a lot of work on solving factorials with induction. –  Tony Oct 21 '13 at 21:58
1  
Hey, that is great. I'll take this and make sure I understand it. Thanks for all of you help. –  Tony Oct 21 '13 at 23:02

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