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

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?

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

using $\left( \begin{array}{c} n \\ 2\ \end{array} \right) = \dfrac{n(n-1)}{2}$?

$$\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = \frac{2n(2n-1)}{2} = 2n(n-1) = 2 \frac{n(n-1)}{2} + n^2 = 2\left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2 $$

Is this proof correct? if not how do I fix it?

Answer 1

It is not correct, and it should go like this:

$$\binom{2n}{2}=\frac{2n(2n-1)}{2}=n(2n-1)=2n^2-n=n(n-1)+n^2=2\frac{n(n-1)}{2}+n^2=2\binom{n}{2}+n^2$$

It looks that your work starts well and ends well but there is a bit of a muddle in the middle.

Answer 2

$$\binom{2n}{2} = \frac{2n(2n-1)}{2} = \frac{2n(n-1) + 2n^2}{2} = 2 \binom{n}{2} + n^2$$

Answer 3

Since OP asks for an algebraic proof, I type this proof making use of number of roots of a degree two polynomials for fun.

Set $f(n) = \left( \begin{array}{c} 2n \\ 2\ \end{array} \right) - 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) - n^2 $. Since $\deg f \le 2$, $f$ has at most two distinct roots. \begin{align} f(2) &= \left( \begin{array}{c} 4 \\ 2\ \end{array} \right) - 2 \left( \begin{array}{c} 2 \\ 2\ \end{array} \right) - 4 = 6-2-4=0 \\ f(3) &= \left( \begin{array}{c} 6 \\ 2\ \end{array} \right) - 2 \left( \begin{array}{c} 3 \\ 2\ \end{array} \right) - 9 = 15-6-9=0 \\ f(4) &= \left( \begin{array}{c} 8 \\ 2\ \end{array} \right) - 2 \left( \begin{array}{c} 4 \\ 2\ \end{array} \right) - 16 = 28-12-16=0 \end{align}

Therefore, $f$ is identically zero. Hence $\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2$.