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

Question

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?

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

Answer 1

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 $n\cdot n=n^2$

Answer 2

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.