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My teacher showed us a proof by induction for this equation for $n\in\mathbb{N}$:

$$\sum\limits_{k=0}^n{{n}\choose{k}} = 2^n$$

In the first step, this sum is rewritten using ${{n+1}\choose{k}}={{n}\choose{k-1}}+{{n}\choose{k}}$.

However, he doesn't explain why this would be - and since he just introduced binomials coefficients, I assume it's something trivial, which I just don't see. I can't figure out why this would hold though.

I tried rewriting the binomial coefficients with ${{n}\choose{k}}=\frac{n!}{k!\cdot(n-k)!}$ when $n,k\in\mathbb{N}$ and $n\geq k$:

$${{n+1}\choose{k}}=\frac{(n+1)!}{k!\cdot(n+1-k)!}$$

$${{n}\choose{k-1}}=\frac{n!}{(k-1)!\cdot(n-k+1)!}$$

$${{n}\choose{k}}=\frac{n!}{k!\cdot(n-k)!}$$

But I can't prove:

$$\frac{(n+1)!}{k!\cdot(n+1-k)!} \stackrel{?}{=} \frac{n!}{(k-1)!\cdot(n-k+1)!} + \frac{n!}{k!\cdot(n-k)!}$$

Am I on the right track? Is there a basic idea behind this equation which makes you see it easily?

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    You are on the right track, but you have some mistakes in your equations. The first one is correct, but then the second should have $n-(k-1)$ instead of $(n-k)$, and the third should have $(n-k)$ instead of $(n+1-k)$. Then you should be able to succeed with showing your big equation is correct. – Matt Nov 20 '14 at 12:25
  • Just so you know this trick, but one can use the binomial formula to get this result instantly: $ (a+b)^n = \sum_{k=0}^n {{n}\choose{k}}a^kb^{n-k} $; you only have to set a=1; b=1 – mvggz Nov 20 '14 at 12:28
  • @Matt thank you, you're correct. Actually I had this on my paper, I just typed it incorrectly. However, that doesn't change the big equation, right? It just swaps the two terms in the right part. (I corrected the mistake in my question) –  Nov 20 '14 at 12:29
  • Thanks @mvggz, I'll note that down, but of course I still want to understand the proof :) –  Nov 20 '14 at 12:31
  • @CamilStaps How does your teacher define the binomial coefficients? You can obviously define it in terms of a fraction of factions, but it's possible to define them recursively based on Pascal's rule. – David H Nov 20 '14 at 12:34
  • @DavidH with ${{n}\choose{k}}\stackrel{\text{def}}{=}\frac{n(n-1(n-2)\dots(n-k+1)}{k!}$, and with the simpler form when $n,k\in\mathbb{N}$ and $k\geq n$ as mentioned in my question. –  Nov 20 '14 at 12:36
  • @Camil Staps Ok, look also at my answer if you've never seen this argument. It's much clearer in my view, because it's almost intuitive :) – mvggz Nov 20 '14 at 12:39
  • In order to solve the equation, try to first get a common denominator using $k! = k (k-1)!$ and $(n-k+1)! = (n-k+1) (n-k)!$... – Klaus Draeger Nov 20 '14 at 12:42
  • @Camil Ok, now the trick to proving this equation is to do things like multiply the numerator and denominator (of the first fraction on the right) by $k$, which will change the $(k-1)!$ in the denominator into $k!$. Similarly, multiply the last fraction by $(n-k+1)$ on the top and bottom, so all three fractions have the same denominator. Then you can add those two fractions together. – Matt Nov 20 '14 at 12:45
  • @Matt great, thanks, you're the first (before the two answers) managing to explain this with algebra to me. I think you should write it as an answer, so that I can upvote it ;) (oh, I see now, with your explanation, that this is the exact same approach as Timbuc's answer uses, too bad) –  Nov 20 '14 at 12:50
  • @Camil Really you were 90% of the way there already on your own! You had already figured out exactly how to approach the problem. I think all that was missing was familiarity with what you can do with fractions involving factorials. Yes, Timbuc's answer explains that remaining step nicely. – Matt Nov 20 '14 at 12:56

4 Answers4

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Algebra: note that $\;r!=r(r-1)!\;,\;\;\frac{r!}r=(r-1)!\;$ , and also correcting your equalities we have

$$\binom n{k-1}=\frac{n!}{(k-1)!(n-k+1)!}\;,\;\;\binom nk=\frac{n!}{k!(n-k)!}$$

so that we actually should have on the RHS of your last equality

$$\frac{n!}{(k-1)!\cdot(n-k+1)!} + \frac{n!}{k!\cdot(n-k)!}=n!\frac{k+n-k+1}{k!(n-k+1)!}=$$

$$=\frac{n!(n+1)}{k!(n-k+1)!}=\binom{n+1}{k}$$

Timbuc
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I can give you an other way of proving this without any calculus.

What is ${{n}\choose{k}}$? This is the number of ways to choose k elements among n ones, without specifying an order.

So let's have the set: A=$(a_1...a_n)$ n elements ( tennis balls per say). Let's look at one particularly: $a_i$ because it is the only red ball of the set.

The number of ways to choose k elements in A is the number of ways to choose k elements with $a_i$ in it, plus the number of ways to choose k elements without $a_i$.

For the first, since you know that among the k elements $a_i$ has to be present you can choose k-1 elements among n-1 elements ($a_i$ has already been chosen so your set is diminished of one ball, that gives n-1 elements left). This gives ${{n-1}\choose{k-1}}$

Likewise, for the second you know that $a_i$ is not part of your choosing so you have to choose k elements among n-1 elements, that is ${{n-1}\choose{k}}$

So you get the equality : ${{n}\choose{k}} = {{n-1}\choose{k-1}} + {{n-1}\choose{k}}$

mvggz
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  • This is wonderful, thank you very much. This may very well end up as the accepted answer, but I will still try to understand the algebra in Timbuc's answer of course. –  Nov 20 '14 at 12:42
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    @Camil Staps, thanks :) . Of course both ways are important, you should understand them well. But you'll see that often you can get yourself out of trouble with bijections, sheer reformulation of the combinatorics problem. Here I only chose to count while settling one element and the answer became straightforward :) – mvggz Nov 20 '14 at 12:47
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    I think that you may indicate in your proof how this argument leads to a generalization of the original identity OP posed, i.e., $\displaystyle\binom{n+k+1}{k}=\displaystyle\sum_{i=0}^k\binom{n+i}{i}$. I think it will enrich your answer. –  Nov 20 '14 at 12:53
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    @user 170039 , yes of course but I'll leave it to Camil Staps if he's interested :) . One can do this with an induction, once he has the result for p=0 (I think your equality should use different index: k among n+p+1 maybe) – mvggz Nov 20 '14 at 13:01
  • No. It's alright. –  Nov 20 '14 at 13:04
  • @user 170039 my bad – mvggz Nov 20 '14 at 13:10
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$$\dfrac{1}{k}+\dfrac{1}{n-k+1}=\dfrac{n+1}{k(n-k+1)}\\ \left(\dfrac{n!}{(k-1)!(n-k)!}\right)\left(\dfrac{1}{k}+\dfrac{1}{n-k+1}\right)=\left(\dfrac{n!}{(k-1)!(n-k)!}\right)\left(\dfrac{n+1}{k(n-k+1)}\right)\\ \displaystyle\binom{n}{k}+\displaystyle\binom{n}{k-1}=\displaystyle\binom{n+1}{k}$$

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you mismatched:

you must verify

$$\frac{(n+1)!}{k!(n+1-k)!}=\frac{n!}{(k-1)!(n+1-k)!}+\frac{n!}{k!(n-k)!}$$

which is straightforward.

cheers

the right hand side: $$\frac{n!}{(k-1)!(n+1-k)!}+\frac{n!}{k!(n-k)!} = \frac{n!.k+n!.(n+1-k)}{k!(n+1-k)!}=\frac{n!(k+n+1-k)}{k!(n+1-k)!}=\frac{n!(n+1)}{k!(n+1-k)!}=\frac{(n+1)!}{k!(n+1-k)!}$$ = the left hand side.

henry G
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  • Thanks, but I corrected this already in my question, and it doesn't look straightforward to me. –  Nov 20 '14 at 12:38