Difference between revisions of "Combinatorial identity"

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* [[Pascal's Triangle]]
 
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[[Category:Theorems]]

Revision as of 14:13, 4 December 2007

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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

${n \choose k}={n-1\choose k-1}+{n-1\choose k}$

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

See also

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