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

Question

My Approach: I will be making use of $$\tag 1\quad{m+n\choose r} = {m\choose 0}{n \choose r} + {m\choose 1}{n\choose r- 1} + \cdot\cdot\cdot + {m\choose r}{n \choose 0}$$ and $$\tag 2{a \choose b}={a\choose a-b}$$

By (1) $${2n\choose n} = {n+n\choose n}= {n\choose 0}{n\choose n}+{n\choose 1}{n\choose n-1}+\cdot\cdot\cdot + {n\choose n}{n\choose 0}$$

By (2) $${2n\choose n} = {n\choose 0}^2 + {n\choose 1}^2 + \cdot\cdot\cdot + {n\choose n}^2$$

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

Please have a look at my solution and give me any hints and\or suggestions you may have.

That's fine. If you already have (1), it is the more general formula. — Thomas Andrews, Sep 23 '14 at 14:33
Other than a typo in the formula following "By (1)..." it is fine (you jump straight to $r = 2$ instead of using $r = 1$ first). — Simon Rose, Sep 23 '14 at 14:51
Kermit: I believe that Thomas means you can derive the initial formula you're trying to prove from (1) (as you've just shown, of course), so (1) is more general than the specific sum for $2n\choose n$. — Steven Stadnicki, Sep 23 '14 at 14:54
@EwanDelanoy You're right, my mistake. I'm not familiar with the procedure, should I delete this question? — Kermit the Hermit, Sep 23 '14 at 15:05
@KermittheHermit There's a simpler way to format the question: have a look at the \tag function in LaTeX. — Théophile, Sep 23 '14 at 15:49
@Théophile Great, that's neat. My formatting was kind of embarrassing. What exactly does \quad do? — Kermit the Hermit, Sep 23 '14 at 19:58
@StevenStadnicki Makes sense, I see now what was meant by Thomas Andrews. — Kermit the Hermit, Sep 23 '14 at 19:59
@KermittheHermit Oops, I didn't mean to leave the \quad in there. It's a spacing command, which I was using at first to replace several of your \, before realizing that the simplest way was with \tag. In any case, \quad is useful for such things as x \in S \quad \forall x \in T, which produces the following: $x \in S \quad \forall x \in T$. Here is an excellent reference page for formatting; see item 13 for spacing. Cheers! — Théophile, Sep 23 '14 at 22:33

score 1 · Answer 1 · answered Sep 23 '14 at 15:58

1

Using the fact that $\sum_{k=0}^{l}\binom{n}{k}\binom{m}{l-k}=\binom{n+m}{l}$. Substituting $l,m=n$ and seeing that $\binom{n}{k}=\binom{n}{n-k}$.

Best regards.

answered Sep 23 '14 at 15:58

Charles Seife

563

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

1 Answers1