prove that $ \binom n 2 + \binom {n-2} 2 + \binom {n-4} 2 + \dots + \binom 3 2 = \frac 1 {24} (n-1)(n+1) (2n+3) $ $$ \binom n 2 + \binom {n-2} 2 + \binom {n-4} 2 + \dots + \binom 3 2 = \frac 1 {24} (n-1)(n+1) (2n+3)  $$ where n is odd.
Plesase help mi with that equation.

Let $n=2m+1\iff m=(n-1)/2$. Then, we have
$$\begin{align}\sum_{k=1}^{m}\binom{2k+1}{2}&=\sum_{k=1}^{m}\frac{(2k+1)(2k)}{2}\\&=\sum_{k=1}^{m}(2k^2+k)\\&=2\cdot\frac{m(m+1)(2m+1)}{6}+\frac{m(m+1)}{2}\\&=\frac{m(m+1)(4m+5)}{6}\\&=\frac 16\times\frac{n-1}{2}\times \left(\frac{n-1}{2}+1\right)\times \left(4\cdot \frac{n-1}{2}+5\right)\\&=\frac{1}{24}(n-1)(n+1)(2n+3).\end{align}$$
