We can prove many relationships between the binomial coefficients by taking out factors and equating coefficients.
Suppose we state with the expression  {jatex options:inline}(1+x)^n{/jatex}.
{jatex options:inline}(1+x)^n=(1+x)^2(1+x)^{n-2} =(1+2x+x^2)(1+x)^{n-2}{/jatex}
{jatex options:inline}\sum_{r=0}^n n \begin{pmatrix}n\\r\end{pmatrix}x^r=(1+2x+x^2) \sum_{r=0}^{n-2} \begin{pmatrix}n-2\\r\end{pmatrix} x^r{/jatex}
{jatex options:inline}\sum_{r=0}^n \begin{pmatrix}n\\r\end{pmatrix}x^r= \sum_{r=0}^{n-2} \begin{pmatrix}n-2\\r\end{pmatrix}x^r + 2 \sum_{r=0}^{n-2} \begin{pmatrix}n-2\\r\end{pmatrix}x^{r+1} + \sum_{r=0}^{n-2} \begin{pmatrix}n-2\\r\end{pmatrix}x^{r+2} {/jatex}
{jatex options:inline}\sum_{r=0}^n \begin{pmatrix}n\\r\end{pmatrix}x^r= \sum_{r=0}^{n-2} \begin{pmatrix}n-2\\r\end{pmatrix}x^r + 2 \sum_{r=1}^{n-1} \begin{pmatrix}n-2\\r-1\end{pmatrix}x^r + \sum_{r=2}^n \begin{pmatrix}n-2\\r-2\end{pmatrix}x^r {/jatex}
Now equating coefficients.
{jatex options:inline}r=0{/jatex}: Both sides equal 1 (last two terms are zero).
{jatex options:inline}r=1{/jatex}:  {jatex options:inline}n=(n-2)+2(1){/jatex}  (last term is zero).
{jatex options:inline}r \ge 2{/jatex}:  {jatex options:inline}\begin{pmatrix}n\\r\end{pmatrix}= \begin{pmatrix}n-2\\r\end{pmatrix}+ 2 \begin{pmatrix}n-2\\r-1\end{pmatrix} + \begin{pmatrix}n-2\\r-2\end{pmatrix} {/jatex}