Hockey-stick Identity

In combinatorial mathematics, the identity

: $\sum^n_= \qquad \text n,r\in\mathbb, \quad n\geq r$

or equivalently, the mirror-image by the substitution $j\to i-r$:

: $\sum^_=\sum^_= \qquad \text n,r\in\mathbb, \quad n\geq r$

is known as the hockey-stick, Christmas stocking identity, boomerang identity, or Chu's Theorem. The name stems from the graphical representation of the identity on Pascal's triangle: when the addends represented in the summation and the sum itself are highlighted, the shape revealed is vaguely reminiscent of those objects (see hockey stick, Christmas stocking).

Proofs

Generating function proof

We have

:$X^r + X^ + \dots + X^ = \frac$

Let $X=1+x$, and compare coefficients of $x^r$.

Inductive and algebraic proofs

The inductive and algebraic proofs both make use of Pascal's identity:

:$=+.$

Inductive proof

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

Base case
Let $n=r$;

:$\sum^n_ = \sum^r_= = 1 = = .$

Inductive step
Suppose, for some $k\in\mathbb, k \geqslant r$,

:$\sum^k_=$

Then

:$\sum^_ = \left(\sum^k_ \right) + =+=.$

Algebraic proof

We use a telescoping argument to simplify the computation of the sum:

:
\begin
\sum_^n \binom
=\sum_^n\binom tk
&= \sum_^n\left \binom -\binom \right\
&=\sum_^\binom - \sum_^n \binom t\\
&=\sum_^\binom - \sum_^n \binom t\\
&=\binom-\underbrace_0&&\text\\
&=\binom.
\end

Combinatorial proofs

Proof 1

Imagine that we are distributing $n$ indistinguishable candies to $k$ distinguishable children. By a direct application of the stars and bars method, there are

:$\binom$

ways to do this. Alternatively, we can first give $0\leqslant i\leqslant n$ candies to the oldest child so that we are essentially giving $n-i$ candies to $k-1$ kids and again, with stars and bars and double counting, we have

:$\binom=\sum_^n\binom,$

which simplifies to the desired result by taking $n' = n+k-2$ and $r=k-2$, and noticing that $n'-n = k-2=r$:

:$\binom=\sum_^n \binom r = \sum_^ \binom r .$

Proof 2

We can form a committee of size $k+1$ from a group of $n+1$ people in

:$\binom$

ways. Now we hand out the numbers $1,2,3,\dots,n-k+1$ to $n-k+1$ of the $n+1$ people. We can divide this into $n-k+1$ disjoint cases. In general, in case $x$, $1\leqslant x\leqslant n-k+1$, person $x$ is on the committee and persons $1,2,3,\dots, x-1$ are not on the committee. This can be done in

:$\binom$

ways. Now we can sum the values of these $n-k+1$ disjoint cases, getting

:$\binom = \binom n k + \binom k + \binom k + \cdots + \binom k+ \binom k k.$

See also

* Pascal's identity
* Pascal's triangle
* Vandermonde's identity

References

{{reflist

External links

On AOPS

On StackExchange, Mathematics

''Pascal's Ladder'' on the Dyalog Chat Forum

Theorems in combinatorics
Mathematical identities
Articles containing proofs
Factorial and binomial topics