## Method For Determining Relationship Between Binomial Coefficients

We can prove many relationships between the binomial coefficients by taking out factors and equating coefficients.
Suppose we state with the expression
$(1+x)^n$
.
$(1+x)^n=(1+x)^2(1+x)^{n-2} =(1+2x+x^2)(1+x)^{n-2}$

$\sum_{r=0}^n n \begin{pmatrix}n\\r\end{pmatrix}x^r=(1+2x+x^2) \sum_{r=0}^{n-2} \begin{pmatrix}n-2\\r\end{pmatrix} x^r$

$\sum_{r=0}^n \begin{pmatrix}n\\r\end{pmatrix}x^r= \sum_{r=0}^{n-2} \begin{pmatrix}n-2\\r\end{pmatrix}x^r + 2 \sum_{r=0}^{n-2} \begin{pmatrix}n-2\\r\end{pmatrix}x^{r+1} + \sum_{r=0}^{n-2} \begin{pmatrix}n-2\\r\end{pmatrix}x^{r+2}$

$\sum_{r=0}^n \begin{pmatrix}n\\r\end{pmatrix}x^r= \sum_{r=0}^{n-2} \begin{pmatrix}n-2\\r\end{pmatrix}x^r + 2 \sum_{r=1}^{n-1} \begin{pmatrix}n-2\\r-1\end{pmatrix}x^r + \sum_{r=2}^n \begin{pmatrix}n-2\\r-2\end{pmatrix}x^r$

Now equating coefficients.
$r=0$
: Both sides equal 1 (last two terms are zero).
$r=1$
:
$n=(n-2)+2(1)$
(last term is zero).
$r \ge 2$
:
$\begin{pmatrix}n\\r\end{pmatrix}= \begin{pmatrix}n-2\\r\end{pmatrix}+ 2 \begin{pmatrix}n-2\\r-1\end{pmatrix} + \begin{pmatrix}n-2\\r-2\end{pmatrix}$

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