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Question Number 185659 by mr W last updated on 25/Jan/23
prove for r, n ∈ N  Σ_(k=r) ^n  ((k),(r) ) = (((n+1)),((r+1)) )  (Hockey−stick identity)
$${prove}\:{for}\:{r},\:{n}\:\in\:\mathbb{N} \\ $$$$\underset{{k}={r}} {\overset{{n}} {\sum}}\begin{pmatrix}{{k}}\\{{r}}\end{pmatrix}\:=\begin{pmatrix}{{n}+\mathrm{1}}\\{{r}+\mathrm{1}}\end{pmatrix} \\ $$$$\left({Hockey}−{stick}\:{identity}\right) \\ $$
Commented by mr W last updated on 25/Jan/23
Answered by mr W last updated on 25/Jan/23
 (((n+1)),((r+1)) )   = ((n),(r) )+ ((n),((r+1)) )  = ((n),(r) )+ (((n−1)),(r) )+ (((n−1)),((r+1)) )  = ((n),(r) )+ (((n−1)),(r) )+ (((n−2)),(r) )+ (((n−2)),((r+1)) )  ......  = ((n),(r) )+ (((n−1)),(r) )+ (((n−2)),(r) )+...+ (((r+1)),(r) )+ (((r+1)),((r+1)) )  = ((n),(r) )+ (((n−1)),(r) )+ (((n−2)),(r) )+...+ (((r+1)),(r) )+ ((r),(r) )  =Σ_(k=r) ^n  ((k),(r) )
$$\begin{pmatrix}{{n}+\mathrm{1}}\\{{r}+\mathrm{1}}\end{pmatrix}\: \\ $$$$=\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{{r}+\mathrm{1}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}+\begin{pmatrix}{{n}−\mathrm{1}}\\{{r}}\end{pmatrix}+\begin{pmatrix}{{n}−\mathrm{1}}\\{{r}+\mathrm{1}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}+\begin{pmatrix}{{n}−\mathrm{1}}\\{{r}}\end{pmatrix}+\begin{pmatrix}{{n}−\mathrm{2}}\\{{r}}\end{pmatrix}+\begin{pmatrix}{{n}−\mathrm{2}}\\{{r}+\mathrm{1}}\end{pmatrix} \\ $$$$…… \\ $$$$=\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}+\begin{pmatrix}{{n}−\mathrm{1}}\\{{r}}\end{pmatrix}+\begin{pmatrix}{{n}−\mathrm{2}}\\{{r}}\end{pmatrix}+…+\begin{pmatrix}{{r}+\mathrm{1}}\\{{r}}\end{pmatrix}+\begin{pmatrix}{{r}+\mathrm{1}}\\{{r}+\mathrm{1}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}+\begin{pmatrix}{{n}−\mathrm{1}}\\{{r}}\end{pmatrix}+\begin{pmatrix}{{n}−\mathrm{2}}\\{{r}}\end{pmatrix}+…+\begin{pmatrix}{{r}+\mathrm{1}}\\{{r}}\end{pmatrix}+\begin{pmatrix}{{r}}\\{{r}}\end{pmatrix} \\ $$$$=\underset{{k}={r}} {\overset{{n}} {\sum}}\begin{pmatrix}{{k}}\\{{r}}\end{pmatrix} \\ $$
Commented by cortano1 last updated on 25/Jan/23
yes...great
$${yes}…{great} \\ $$

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