I came across the following proof in my textbook that was used as a end of chapter review. How can I prove the following algebraically?

$$\left( \begin{array}{c} 2n \\ n\ \end{array} \right) = 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2$$


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  • 6
    $\begingroup$ Is there a typo? I don't think that's true $\endgroup$ – John Doe Feb 2 '18 at 15:25
  • $\begingroup$ n=1 is false?! Isn't it? $\endgroup$ – Ivan Di Liberti Feb 2 '18 at 15:28
  • 2
    $\begingroup$ Left hand side should be $\left( \begin{array}{c} 2n \\ 2\ \end{array} \right)$ I think. $\endgroup$ – Marc Paul Feb 2 '18 at 15:29
  • $\begingroup$ FWIW, $\binom{2n}{n}\geq \frac{4^{n}}{2n+1},$ so if that is what the book said, it is way off base because the left side increases exponentially, while the right side does not. $\endgroup$ – Thomas Andrews Feb 2 '18 at 15:38
  • $\begingroup$ yes sorry that was a typo! $\endgroup$ – user104 Feb 2 '18 at 16:19

As an alternative, with the typo correction we have

$$\begin{eqnarray*} \binom{2n}{2} = 2 \binom{n}{2}+n^2 \end{eqnarray*}$$

which can be interpreted in this way

  • LHS is the number of couples we can choose among $2n$
  • RHS is the same choice mabe by dividing in two groups of n; indeed we can choose $\binom{n}{2}$ couples from the first group + $\binom{n}{2}$ couples from the second group and then the couples made of 1 from the first group and 1 from the second that is $n\cdot n=n^2$
  • $\begingroup$ is this a complete proof? $\endgroup$ – user104 Feb 2 '18 at 16:20
  • $\begingroup$ would this be considered a algebraic proof or combinatorial? $\endgroup$ – user104 Feb 2 '18 at 16:22
  • 1
    $\begingroup$ It is combinatorial of course, I’ve given it as an interesting alternative $\endgroup$ – gimusi Feb 2 '18 at 16:30
  • $\begingroup$ okay this makes sense as a complete combinatorial proof, thanks! $\endgroup$ – user104 Feb 2 '18 at 16:36
  • $\begingroup$ Maybe of interest:math.stackexchange.com/questions/2533960/… $\endgroup$ – Peter Szilas Feb 2 '18 at 16:45

Typo alert \begin{eqnarray*} \binom{2n}{\color{red}{2}} = 2 \binom{n}{2}+n^2. \end{eqnarray*}

Now use $\binom{n}{2} =\frac{n(n-1)}{2}$ and the above formula is easy to prove.

  • $\begingroup$ thanks that was a typo, could you show me how to use that formula? I haven't seen it before $\endgroup$ – user104 Feb 2 '18 at 16:20
  • $\begingroup$ would this be considered a algebraic proof or combinatorial? $\endgroup$ – user104 Feb 2 '18 at 16:22
  • $\begingroup$ Yes, this is an algebriac proof ... still trying to think of a nice combinatorial arguement that demonstrates the formula ... $\endgroup$ – Donald Splutterwit Feb 2 '18 at 16:32
  • $\begingroup$ how do I use that formula you gave? that would conclude the algebraic proof $\endgroup$ – user104 Feb 2 '18 at 16:36
  • $\begingroup$ \begin{eqnarray*} \binom{2n}{2} = \frac{ 2n(2n-1)}{2} = \cdots \end{eqnarray*} and now see DrSG's solution. $\endgroup$ – Donald Splutterwit Feb 2 '18 at 16:39

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