Hockey-stick identity explained

In combinatorial mathematics, the hockey-stick identity,[1] Christmas stocking identity,[2] boomerang identity, Fermat's identity or Chu's Theorem,[3] states that if

n\geqr\ge0

are integers, then

\binom{r}{r}+\binom{r+1}{r}+\binom{r+2}{r}++\binom{n}{r}=\binom{n+1}{r+1}.

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).

Formulations

Using sigma notation, the identity states

n
\sum
i=r

{i\chooser}={n+1\chooser+1}    forn,r\inN,n\geqr

or equivalently, the mirror-image by the substitution

j\toi-r

:
n-r
\sum
j=0

{j+r\choose

n-r
r}=\sum
j=0

{j+r\choosej}={n+1\choosen-r}    forn,r\inN,n\geqr.

Proofs

Generating function proof

Let

X=1+x

. Then, by the partial sum formula for geometric series, we find that

Xr+Xr+1+...+Xn=

Xr-Xn+1
1-X

=

Xn+1-Xr
x
.Further, by the binomial theorem, we also find that

Xr=(1+x)r=

r+k
\sum
i=0

\binom{r+k}{i}xi

.

Note that this means the coefficient of

xr

in

Xr

is given by

\binom{r+k}{r}

.

Thus, the coefficient of

xr

in the left hand side of our first equation can be obtained by summing over the coefficients of

xr

from each term, which gives
n-r
\sum
k=0

\binom{r+k}{r}

Similarly, we find that the coefficient of

xr

on the right hand side is given by the coefficient of

xr

in

Xn+1-Xr

, which is

\binom{n+1}{r+1}-\binom{r}{r+1}=\binom{n+1}{r+1}

Therefore, we can compare the coefficients of

xr

on each side of the equation to find that
n-r
\sum
k=0

\binom{r+k}{r}=\binom{n+1}{r+1}

Inductive and algebraic proofs

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

{n\choosek}={n-1\choosek-1}+{n-1\choosek}.

Inductive proof

This identity can be proven by mathematical induction on

n

.

Base caseLet

n=r

;
n
\sum
i=r

{i\chooser}=

r
\sum
i=r

{i\chooser}={r\chooser}=1={r+1\chooser+1}={n+1\chooser+1}.

Inductive stepSuppose, for some

k\inN,k\geqslantr

,
k
\sum
i=r

{i\chooser}={k+1\chooser+1}

Then

k+1
\sum
i=r

{i\chooser}=

k
\left(\sum
i=r

{i\chooser}\right)+{k+1\chooser}={k+1\chooser+1}+{k+1\chooser}={k+2\chooser+1}.

Algebraic proof

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

\begin{align} \sumt=\color{blue0}n\binom{t}{k}=\sumt=\color{bluek}n\binomtk &=

n\left[
\sum
t=k

\binom{t+1}{k+1}-\binom{t}{k+1}\right]\\ &=\sumt=\color{greenk}\color{greenn}\binom{\color{green}{t+1}}{k+1}-

n
\sum
t=k

\binomt{k+1}\\ &=\sumt=\color{green{k+1}}\color{green{n+1}}\binom{\color{green}{t}}{k+1}-

n
\sum
t=k

\binomt{k+1}\\ &=\binom{n+1}{k+1}-\underbrace{\binomk{k+1}}0&&bytelescoping\\ &=\binom{n+1}{k+1}. \end{align}

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{n+k-1}{k-1}

ways to do this. Alternatively, we can first give

0\leqslanti\leqslantn

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{n+k-1}{

n\binom{n+k-2-i}{k-2},
k-1}=\sum
i=0

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{n'+1}{

n
r+1}=\sum
i=0

\binom{n'-i}r=

n'
\sum
i=r

\binom{i}r.

Proof 2

We can form a committee of size

k+1

from a group of

n+1

people in

\binom{n+1}{k+1}

ways. Now we hand out the numbers

1,2,3,...,n-k+1

to

n-k+1

of the

n+1

people. We can then divide our committee-forming process into

n-k+1

exhaustive and disjoint cases based on the committee member with the lowest number,

x

. Note that there are only

k

people without numbers, meaning we must choose at least one person with a number in order to form a committee of

k+1

people. In general, in case

x

, person

x

is on the committee and persons

1,2,3,...,x-1

are not on the committee. The rest of the committee can then be chosen in

\binom{n-x+1}{k}

ways. Now we can sum the values of these

n-k+1

disjoint cases, and using double counting, we obtain

\binom{n+1}{k+1}=\binomnk+\binom{n-1}k+\binom{n-2}k++\binom{k+1}k+\binomkk.

See also

External links

Notes and References

  1. CH Jones (1996) Generalized Hockey Stick Identities and N-Dimensional Block Walking. Fibonacci Quarterly 34(3), 280-288.
  2. Web site: Christmas Stocking Theorem. W.. Weisstein, Eric. mathworld.wolfram.com. en. 2016-11-01.
  3. Book: Merris, Russell . Combinatorics . 2003 . Wiley-Interscience . 0-471-45849-X . 2nd . Hoboken, N.J. . 45 . 53121765.