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

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?

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

closed as off-topic by Andrés E. Caicedo, Did, Hans Engler, hardmath, caveracFeb 2 '18 at 19:28

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• Is there a typo? I don't think that's true – John Doe Feb 2 '18 at 15:25
• n=1 is false?! Isn't it? – Ivan Di Liberti Feb 2 '18 at 15:28
• Left hand side should be $\left( \begin{array}{c} 2n \\ 2\ \end{array} \right)$ I think. – Marc Paul Feb 2 '18 at 15:29
• 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. – Thomas Andrews Feb 2 '18 at 15:38
• yes sorry that was a typo! – user104 Feb 2 '18 at 16:19

2 Answers

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 $n\cdot n=n^2$
• is this a complete proof? – user104 Feb 2 '18 at 16:20
• would this be considered a algebraic proof or combinatorial? – user104 Feb 2 '18 at 16:22
• It is combinatorial of course, I’ve given it as an interesting alternative – gimusi Feb 2 '18 at 16:30
• okay this makes sense as a complete combinatorial proof, thanks! – user104 Feb 2 '18 at 16:36
• Maybe of interest:math.stackexchange.com/questions/2533960/… – 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.

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