Question Number 182437 by akolade last updated on 09/Dec/22
Answered by qaz last updated on 09/Dec/22
^(n+2) =[x^r ]((1+x)^n +2x(1+x)^n +x^2 (1+x)^n ) =[x^r ](1+x)^n +2[x^(r−1) ](1+x)^n +[x^(r−2) ](1+x)^n = ((n),(r) )+2 ((( n)),((r−1)) )+ ((( n)),((r−2)) ) ,by egorychev′s method](https://www.tinkutara.com/question/Q182446.png)
$$\begin{pmatrix}{{n}+\mathrm{2}}\\{\:\:\:\:{r}}\end{pmatrix}=\left[{x}^{{r}} \right]\left(\mathrm{1}+{x}\right)^{{n}+\mathrm{2}} =\left[{x}^{{r}} \right]\left(\left(\mathrm{1}+{x}\right)^{{n}} +\mathrm{2}{x}\left(\mathrm{1}+{x}\right)^{{n}} +{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)^{{n}} \right) \\ $$$$=\left[{x}^{{r}} \right]\left(\mathrm{1}+{x}\right)^{{n}} +\mathrm{2}\left[{x}^{{r}−\mathrm{1}} \right]\left(\mathrm{1}+{x}\right)^{{n}} +\left[{x}^{{r}−\mathrm{2}} \right]\left(\mathrm{1}+{x}\right)^{{n}} \\ $$$$=\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}+\mathrm{2}\begin{pmatrix}{\:\:\:{n}}\\{{r}−\mathrm{1}}\end{pmatrix}+\begin{pmatrix}{\:\:\:{n}}\\{{r}−\mathrm{2}}\end{pmatrix}\:\:\:\:\:,{by}\:{egorychev}'{s}\:{method} \\ $$
Commented by akolade last updated on 17/Dec/22
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$