Difference between revisions of "Combinatorial identity"

(Hockey-Stick Identity)
m (Hockey-Stick Identity)
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.
 
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===
 
===Proof===
 
This identity can be proven by induction on <math>n</math>.
 
This identity can be proven by induction on <math>n</math>.

Revision as of 10:36, 7 September 2006

This article is a stub. Help us out by expanding it.

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.


An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.


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

See also

Invalid username
Login to AoPS