Difference between revisions of "Combinatorial identity"

Revision as of 13:35, 8 December 2007

Hockey-Stick Identity

For $n,r\in\mathbb{N}, n>r,\sum^n_{i=r}{i\choose r}={n+1\choose r+1}$ .

This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself are highlighted, a hockey-stick shape is revealed.

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Proof

This identity can be proven by induction on $n$ .

Base case Let $n=r$ .

$\sum^n_{i=r}{i\choose r}=\sum^r_{i=r}{i\choose r}={r\choose r}=1={r+1\choose r+1}$ .

Inductive step Suppose, for some $k\in\mathbb{N}, k>r$ , $\sum^k_{i=r}{i\choose r}={k+1\choose r+1}$ . Then $\sum^{k+1}_{i=r}{i\choose r}=\left(\sum^k_{i=r}{i\choose r}\right)+{k+1\choose r}={k+1\choose r+1}+{k+1\choose r}={k+2\choose r+1}$ .

It can also be proven algebraicly with pascal's identity

Look at ${r \choose r}+{r+1 \choose r} +{r+2 \choose r}...+{r+a \choose r}$ It can be rewritten as ${r+1 \choose r+1}+{r+1 \choose r} +{r+2 \choose r}...+{r+a \choose r}$ Using pascals identity, we get ${r+2 \choose r+1}+{r+2 \choose r}+...+{r+a \choose r}$ We can continuously apply pascals identity until we get to ${r+a \choose r-1}+{r+a \choose r}={r+a+1 \choose r+1}$

Vandermonde's Identity

Examples

AIME 2000II/7

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 ==Hockey-Stick Identity==
 For <math>n,r\in\mathbb{N}, n>r,\sum^n_{i=r}{i\choose r}={n+1\choose r+1}</math>.
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 [[Category:Theorems]]
+[[Category:Combinatorics]]

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