Difference between revisions of "Combinatorial identity"

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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 $\sum^{k+1}_{i=r}{i\choose r}=\left(\sum^k_{i=r}{i\choose r}\right)+{k+1\choose r}={k+1\choose r+1}+{k+1\choose r}={k+2\choose r+1}$ .

Vandermonde's Identity

Examples

AIME 2000II/7

@@ Line 4: / Line 4: @@
 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.
+{{image}}
 ===Proof===
 This identity can be proven by induction on <math>n</math>.

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