Question Number 74337 by Maclaurin Stickker last updated on 22/Nov/19
$${find}\:{the}\:{contracted}\:{form}\:{of}: \\ $$$$\begin{pmatrix}{{n}}\\{{p}}\end{pmatrix}+\mathrm{2}\begin{pmatrix}{\:\:\:{n}}\\{{p}+\mathrm{1}}\end{pmatrix}+\begin{pmatrix}{\:\:\:{n}}\\{{p}+\mathrm{2}}\end{pmatrix} \\ $$
Answered by MJS last updated on 23/Nov/19
$$\begin{pmatrix}{{n}}\\{{p}}\end{pmatrix}\:=\frac{{n}!}{{p}!\left({n}−{p}\right)!}=\frac{{n}!\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)}{{p}!\left({n}−{p}\right)!\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)} \\ $$$$\mathrm{2}\begin{pmatrix}{{n}}\\{{p}+\mathrm{1}}\end{pmatrix}\:=\frac{\mathrm{2}{n}!}{\left({p}+\mathrm{1}\right)!\left({n}−{p}−\mathrm{1}\right)!}=\frac{\mathrm{2}{n}!\left({n}−{p}\right)}{{p}!\left({n}−{p}\right)!\left({p}+\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{2}{n}!\left({n}−{p}\right)\left({p}+\mathrm{2}\right)}{{p}!\left({n}−{p}\right)!\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)} \\ $$$$\begin{pmatrix}{{n}}\\{{p}+\mathrm{2}}\end{pmatrix}\:=\frac{{n}!}{\left({p}+\mathrm{2}\right)!\left({n}−{p}−\mathrm{2}\right)!}= \\ $$$$=\frac{{n}!\left({n}−{p}\right)\left({n}−{p}−\mathrm{1}\right)}{{p}!\left({n}−{p}\right)!\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)} \\ $$$$ \\ $$$$\begin{pmatrix}{{n}}\\{{p}}\end{pmatrix}\:+\mathrm{2}\begin{pmatrix}{{n}}\\{{p}+\mathrm{1}}\end{pmatrix}\:+\begin{pmatrix}{{n}}\\{{p}+\mathrm{2}}\end{pmatrix}\:= \\ $$$$=\frac{{n}!\left(\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)+\mathrm{2}\left({n}−{p}\right)\left({p}+\mathrm{2}\right)+\left({n}−{p}\right)\left({n}−{p}−\mathrm{1}\right)\right)}{{p}!\left({n}−{p}\right)!\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)}= \\ $$$$=\frac{{n}!\left({n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}\right)}{{p}!\left({n}−{p}\right)!\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)}= \\ $$$$=\frac{{n}!\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{{p}!\left({n}−{p}\right)!\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)}=\frac{\left({n}+\mathrm{2}\right)!}{\left({p}+\mathrm{2}\right)!\left({n}−{p}\right)!}= \\ $$$$=\frac{\left({n}+\mathrm{2}\right)!}{\left({p}+\mathrm{2}\right)!\left(\left({n}+\mathrm{2}\right)−\left({p}+\mathrm{2}\right)\right)!}=\begin{pmatrix}{{n}+\mathrm{2}}\\{{p}+\mathrm{2}}\end{pmatrix} \\ $$