# Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$

After reading this question, the most popular answer use the identity $$\sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1}.$$

What's the name of this identity? Is it the identity of the Pascal's triangle modified.

How can we prove it? I tried by induction, but without success. Can we also prove it algebraically?

EDIT 01 : 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.

• It is sometimes called the "hockey stick". – user940 Oct 21 '15 at 15:24
• There is another cute graphical illustration on the plane of $\binom{n}{k}$ – Eli Korvigo Oct 21 '15 at 16:54
• It's pretty straightforward from the picture. Just switch the $1$ at the top of the stick with the $1$ directly below, then repeatedly replace adjacent numbers with the number in the cell below. This can be translated into a formal proof with words and symbols, but an animation or series of pictures is much more effective. – user2357112 supports Monica Oct 22 '15 at 3:24
• See also this question. Some post which are linked there might be of interest, too. – Martin Sleziak Jan 18 '16 at 15:05
• May I ask where this image is from? I would really like to use it in the Wikipedia article on the Hockey stick identity but of course I want to credit the source – Maximilian Janisch Dec 19 '19 at 19:49

This is purely algebraic. First of all, since $$\dbinom{t}{k} =0$$ when $$k>t$$ we can rewrite the identity in question as $$\binom{n+1}{k+1} = \sum_{t=0}^{n} \binom{t}{k}=\sum_{t=k}^{n} \binom{t}{k}$$

Recall that (by the Pascal's Triangle), $$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$

Hence $$\binom{t+1}{k+1} = \binom{t}{k} + \binom{t}{k+1} \implies \binom{t}{k} = \binom{t+1}{k+1} - \binom{t}{k+1}$$

Let's get this summed by $$t$$: $$\sum_{t=k}^{n} \binom{t}{k} = \sum_{t=k}^{n} \binom{t+1}{k+1} - \sum_{t=k}^{n} \binom{t}{k+1}$$

Let's factor out the last member of the first sum and the first member of the second sum: $$\sum _{t=k}^{n} \binom{t}{k} =\left( \sum_{t=k}^{n-1} \binom{t+1}{k+1} + \binom{n+1}{k+1} \right) -\left( \sum_{t=k+1}^{n} \binom{t}{k+1} + \binom{k}{k+1} \right)$$

Obviously $$\dbinom{k}{k+1} = 0$$, hence we get $$\sum _{t=k}^{n} \binom{t}{k} =\binom{n+1}{k+1} +\sum_{t=k}^{n-1} \binom{t+1}{k+1} -\sum_{t=k+1}^{n} \binom{t}{k+1}$$

Let's introduce $$t'=t-1$$, then if $$t=k+1 \dots n, t'=k \dots n-1$$, hence $$\sum_{t=k}^{n} \binom{t}{k} = \binom{n+1}{k+1} +\sum_{t=k}^{n-1} \binom{t+1}{k+1} -\sum_{t'=k}^{n-1} \binom{t'+1}{k+1}$$

The latter two arguments eliminate each other and you get the desired formulation $$\binom{n+1}{k+1} = \sum_{t=k}^{n} \binom{t}{k} = \sum_{t=0}^{n} \binom{t}{k}$$

Imagine the first $$n + 1$$ numbers, written in order on a piece of paper. The right hand side asks in how many ways you can pick $$k+1$$ of them. In how many ways can you do this?

You first pick a highest number, which you circle. Call it $$s$$. Next, you still have to pick $$k$$ numbers, each less than $$s$$, and there are $$\binom{s - 1}{k}$$ ways to do this.

Since $$s$$ is ranging from $$1$$ to $$n+1$$, $$t:= s-1$$ is ranging from $$0$$ to $$n$$ as desired.

• Why did you call the highest number $s$? Why not call it $s + 1$? Isn't it easier if you call it $s+1$, because there are $\dbinom{s}{k}$ ways to pick $k$ numbers? You don't need to change the bounds or range of summation, as you do in your last para. – koss Jul 11 at 20:19

We can use the well known identity $$1+x+\dots+x^n = \frac{x^{n+1}-1}{x-1}.$$ After substitution $x=1+t$ this becomes $$1+(1+t)+\dots+(1+t)^n=\frac{(1+t)^{n+1}-1}t.$$ Both sides of these equations are polynomials in $t$. (Notice that the RHS simplifies to $\sum_{j=1}^{n+1}\binom {n+1}j t^{j-1}$.)

If we compare coefficient of $t^{k}$ on the LHS and the RHS we see that $$\binom 0k + \binom 1k + \dots + \binom nk = \binom{n+1}{k+1}.$$

This proof is basically the same as the proof using generating functions, which was posted in other answers. However, I think it is phrased a bit differently. (And if it is formulated this way, even somebody who has never heard of generating functions can follow the proof.)

\begin{align} \sum_{t=\color{blue}0}^n \binom{t}{k} =\sum_{t=\color{blue}k}^n\binom tk&= \sum_{t=k}^n\left[ \binom {t+1}{k+1}-\binom {t}{k+1}\right]\\ &=\sum_{t=\color{orange}k}^\color{orange}n\binom {\color{orange}{t+1}}{k+1}-\sum_{t=k}^n\binom t{k+1}\\ &=\sum_{t=\color{orange}{k+1}}^{\color{orange}{n+1}}\binom {\color{orange}{t}}{k+1}-\sum_{t=k}^n\binom t{k+1}\\ &=\binom{n+1}{k+1}-\underbrace{\binom k{k+1}}_0&&\text{by telescoping}\\ &=\binom{n+1}{k+1}\quad\blacksquare\\ \end{align}

You can use induction on $n$, observing that

$$\sum_{t=0}^{n+1} \binom{t}{k} = \sum_{t=0}^{n} \binom{t}{k} + \binom{n+1}{k} = \binom{n+1}{k+1} + \binom{n+1}{k} = \binom{n+2}{k+1}$$

• How can you say that $\sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1}$ in your proof. – hlapointe Oct 21 '15 at 15:13
• That's the inductive hypothesis. – Michael Biro Oct 21 '15 at 15:14
• Ok. Can we prove it algebraically? – hlapointe Oct 21 '15 at 15:15
• What's the first step!? Because if I take $n=1$, the hypothesis seem to be incorrect. – hlapointe Oct 21 '15 at 15:21
• @hlapointe One choice of base case for every fixed $k$ is that $\sum_{t=0}^{k} \binom{t}{k} = \binom{k}{k} = 1 = \binom{k+1}{k+1}$. – Michael Biro Oct 21 '15 at 16:28

Another technique is to use snake oil. Call your sum:

\begin{align} S_k &= \sum_{0 \le t \le n} \binom{t}{k} \end{align}

Define the generating function:

\begin{align} S(z) &= \sum_{k \ge 0} S_k z^k \\ &= \sum_{k \ge 0} z^k \sum_{0 \le t \le n} \binom{t}{k} \\ &= \sum_{0 \le t \le n} \sum_{k \ge 0} \binom{t}{k} z^k \\ &= \sum_{0 \le t \le n} (1 + z)^t \\ &= \frac{(1 + z)^{n + 1} - 1}{(1 + z) - 1} \\ &= z^{-1} \left( (1 + z)^{n + 1} - 1 \right) \end{align}

So we are interested in the coefficient of $z^k$ of this:

\begin{align} [z^k] z^{-1} \left( (1 + z)^{n + 1} - 1 \right) &= [z^{k + 1}] \left( (1 + z)^{n + 1} - 1 \right) \\ &= \binom{n + 1}{k + 1} \end{align}

The RHS is the number of $k+1$ subsets of $\{1,2,...,n+1\}$. Group them according to the largest element in the subset. Sum up all the cases. Get the LHS.

We can use the integral representation of the binomial coefficient $$\dbinom{t}{k}=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{t}}{z^{k+1}}dz\tag{1}$$ and get $$\sum_{t=0}^{n}\dbinom{t}{k}=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\sum_{k=0}^{n}\left(1+z\right)^{t}}{z^{k+1}}dz$$ $$=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(z+1\right)^{n+1}}{z^{k+2}}dz-\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{1}{z^{k+2}}dz$$ and so usign again $(1)$ we have $$\sum_{t=0}^{n}\dbinom{t}{k}=\dbinom{n+1}{k+1}-0=\color{red}{\dbinom{n+1}{k+1}.}$$

• It is so nice and weird. +1 – Behrouz Maleki Jul 5 '16 at 10:27
• +1. Nice work. You must subtract $\displaystyle{\delta_{k,-1}}$ in order to take account of the case $\displaystyle{k = -1}$. When $\displaystyle{k = -1}$, the LHS is equal to $\displaystyle{0}$ and your RHS is equal to $\displaystyle{1}$. With the $\displaystyle{\delta_{k,-1}}$ you'll get $\displaystyle{1 - 1 = 0}$. – Felix Marin Jul 6 '16 at 21:50

A Generalization

In this answer, I prove the identity $$\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}\tag{1}$$ Here is a generalization of the identity in question, proven using the Vandermonde Identity \begin{align} \sum_{t=0}^n\binom{t}{k}\binom{n-t}{j} &=\sum_{t=0}^n\binom{t}{t-k}\binom{n-t}{n-t-j}\tag{2}\\ &=\sum_{t=0}^n(-1)^{t-k}\binom{-k-1}{t-k}(-1)^{n-t-j}\binom{-j-1}{n-t-j}\tag{3}\\ &=(-1)^{n-j-k}\sum_{t=0}^n\binom{-k-1}{t-k}\binom{-j-1}{n-t-j}\tag{4}\\ &=(-1)^{n-j-k}\binom{-k-j-2}{n-j-k}\tag{5}\\ &=\binom{n+1}{n-j-k}\tag{6}\\ &=\binom{n+1}{j+k+1}\tag{7} \end{align} Explanation:
$$(2)$$: $$\binom{n}{k}=\binom{n}{n-k}$$
$$(3)$$: apply $$(1)$$ to each binomial coefficient
$$(4)$$: combine the powers of $$-1$$ which can then be pulled out front
$$(5)$$: apply Vandermonde
$$(6)$$: apply $$(1)$$
$$(7)$$: $$\binom{n}{k}=\binom{n}{n-k}$$

To get the identity in the question, set $$j=0$$.

A Simpler Proof of the Basic Formula \begin{align} \sum_{k=0}^n\color{#C00}{\binom{k}{m}} &=\sum_{k=0}^n\color{#C00}{\left[x^m\right](1+x)^k}\tag8\\ &=\left[x^m\right]\frac{(1+x)^{n+1}-1}{(1+x)-1}\tag9\\[6pt] &=\left[x^{m+1}\right](1+x)^{n+1}-1\tag{10}\\[6pt] &=\binom{n+1}{m+1}\tag{11} \end{align} Explanation:
$$\phantom{1}\text{(8)}$$: use the definition of the binomial coefficient
$$\phantom{1}\text{(9)}$$: sum the finite geometric series
$$(10)$$: multiply by $$x$$ and adjust the exponent of $$x$$
$$(11)$$: extract the coefficient using the binomial theorem

• @FoF: I have added a link here and answered your other question. Thanks for mentioning the difficulty. – robjohn Dec 7 '13 at 12:33
• @FoF: That is the Vandermonde Identity that I mentioned at the beginning. – robjohn Dec 8 '13 at 18:56
• @FoF: I added an explanation for each line. – robjohn Dec 9 '13 at 2:20
• I answered my own question about $(5, 6$) here. – NaN Dec 10 '13 at 8:54
• @FoF: Ah. That is why I added the Explanation when I saw difficulty in following the argument. – robjohn Dec 11 '13 at 7:46

You remember that: $$(1+x)^m = \sum_k \binom{m}{k} x^k$$ So the sum $$\sum_{m=0}^M \binom{m+k}{k}$$ is the coefficient of $x^k$ in: $$\sum_{m=0}^M (1+x)^{m+k}$$ Yes? So now use the geometric series formula given: $$\sum_{m=0}^M (1+x)^{m+k} = -\frac{(1+x)^k}{x} \left( 1 - (1+x)^{M+1} \right)$$ And now you want to know what is coefficient of $x^k$ in there. You got it from here.

Recall that for $k\in\Bbb N$ we have the generating function

$$\sum_{n\ge 0}\binom{n+k}kx^n=\frac1{(1-x)^{k+1}}\;.$$

The identity in the question can therefore be rewritten as

$$\left(\sum_{n\ge 0}\binom{n+k}kx^n\right)\left(\sum_{n\ge 0}x^n\right)=\sum_{n\ge 0}\binom{n+k+1}{k+1}x^n\;.$$

The coefficient of $x^n$ in the product on the left is

$$\sum_{i=0}^n\binom{i+k}k\cdot1=\sum_{i=0}^n\binom{i+k}k\;,$$

and the $n$-th term of the discrete convolution of the sequences $\left\langle\binom{n+k}k:n\in\Bbb N\right\rangle$ and $\langle 1,1,1,\dots\rangle$. And at this point you’re practically done.

• Is there a typo in the second equation (first sum)? I believe $k$ should be indexed. – TheRealFakeNews May 27 '13 at 6:20
• @Alan: No, the sum is over $n$; $k$ is fixed throughout. – Brian M. Scott May 27 '13 at 7:19
• In my text, I have an identity $\sum_{r\geq 0} \binom{r + n}{r} x^r = 1/(1-x)^{n+1}$ This may be the cause of my confusion, but is this identity correct and is it equivalent to the one you used? – TheRealFakeNews May 27 '13 at 8:22
• @Alan: Sure: your $r$ is my $n$, and your $n$ is my $k$. – Brian M. Scott May 27 '13 at 8:28
• @Alan: $\binom{r+n}r=\binom{r+n}n$; now do the translation. (Sorry: I didn’t notice before that you’d used the symmetrically opposite binomial coefficient.) – Brian M. Scott May 27 '13 at 19:19

We can prove this by counting in two ways.

Let $$S$$ be the set of all $$(k+1)$$-element subsets of $$[n+1]$$. By definition, $$|S|=\binom{n+1}{k+1}$$.

Let $$S_i$$ be the set of all $$(k+1)$$-element subsets of $$[n+1]$$ such that the largest element is $$i+1$$. Picking $$k+1$$ elements from $$[n+1]$$ such that the largest element is $$i+1$$ is a two-step-process.

(Step 1) Pick $$i+1$$. The number of way(s) to do this is $$\binom{1}{1}$$.

(Step 2) Pick the $$k$$ elements from the the remaining $$i$$ elements. The number of way(s) to do this is $$\binom{i}{k}$$.

Therefore, $$|S_i|=\binom{1}{1}\binom{i}{k}=\binom{i}{k}$$. Since we can see that $$S_k, S_{k+1}, S_{k+2}, \dots, S_n$$ partition $$S$$, we have that $$\begin{gather*} \sum_{i=k}^n|S_i|=|S|\\ \sum_{i=k}^n\binom{i}{k}=\binom{n+1}{k+1} \end{gather*}$$ Since we know that if $$i < k$$, then $$\binom{i}{k}=0$$, we can say that $$\sum_{i=k}^n\binom{i}{k}=\sum_{i=0}^n\binom{i}{k}$$. Therefore, we have $$\begin{gather*} \sum_{i=0}^n \binom{i}{k} = \binom{n+1}{k+1} \end{gather*}$$

A standard technique to prove such identities $$\sum_{i=0}^Mf(i)=F(M)$$, involving on one hand a sum where only the upper bound $$M$$ is variable and on the other hand an explicit expression in terms of$$~M$$, is to use induction on$$~M$$. It amounts to showing that $$f(M)=F(M)-F(M-1)$$ (and that $$F(0)=f(0)$$). This is similar to using the fundamental theorem of calculus in showing that $$\int_0^{x_0}f(x)\mathrm dx=F(x_0)$$ by establishing $$f(x)=F'(x)$$ (and $$F(0)=0$$).

So here you need to check (apart from the obvious starting case $$M=0$$) that $$\binom{M+k}k=\binom{M+k+1}{k+1}-\binom{M+k}{k+1}$$. This is just in instance of Pascal's recurrence for binomial coefficients.

Added remark this identity (not its name) is very old. It is one of the first "conséquences" that Pascal gives in his "Traité du Triangle arithmétique" after defining this triangle by means of (what is now called) Pascal's recursion. Indeed, it is either the "conséquence seconde" or the "conséquence troisième", depending on how one identifies the Triangle arithmétique which is a rectangular table with modern depictions of the triangle: if one has the "columns" (rangs perpendiculaires) correspond to sets of $$\binom nk$$ with $$k$$ fixed, while "rows" (rangs parallèles) correspond to sets of $$\binom nk$$ with $$n-k$$ fixed (this is geometrically most natural, basically a rotation by $$-\frac\pi4$$), then it is the "conséquence troisième", but if one respects the combinatorial interpretation Pascal gives (much later) in Proposition II, then identification differs by a symmetry of the triangle, and one gets the "conséquence seconde", which talks about sums along rows rather than columns. (For comparison, the "conséquence première" that every entry on the border of the triangle equals$$~1$$.)

CONSÉQUENCE TROISIÈME. En tout Triangle arithmétique, chaque cellule égale la somme de toutes celles du rang perpendiculaire précédent, comprises depuis son rang parallèle jusqu'au premier inclusivement.

Loosely translated: in every Pascal's triangle, each entry equals the sum of those of the previous column, from that of its (own) row up to the first (row) inclusive.

The terms $$\binom tk$$ count the ways to distribute $$t-k$$ balls over $$k+1$$ bins, and we want to show that they sum to $$\binom{n+1}{k+1}$$, the number of ways to distribute $$n-k$$ balls over $$k+2$$ bins. Designate one of the $$k+2$$ bins as special and enumerate the ways to distribute $$n-k$$ balls over the $$k+2$$ bins according to the number $$n-t$$ of balls placed in the designated bin, with the remaining $$t-k$$ balls distributed over the remaining $$k+1$$ bins.

$$\newcommand{\angles}{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}{\overline{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$$ Assuming $$\ds{M \geq 0}$$:

\begin{align} & \mbox{Note that} \\[2mm] &\ \sum_{m = 0}^{M}{m + k \choose k} = \sum_{m = k}^{M + k}{m \choose k} = a_{M + k} - a_{k - 1}\quad\mbox{where}\quad a_{n} \equiv \sum_{m = 0}^{n}{m \choose k}\tag{1} \end{align}

Then, \begin{align} \color{#f00}{a_{n}} & \equiv \sum_{m = 0}^{n}{m \choose k} = \sum_{m = 0}^{n}\ \overbrace{% \oint_{\verts{z} = 1}{\pars{1 + z}^{m} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}} ^{\ds{m \choose k}}\ =\ \oint_{\verts{z} = 1}{1 \over z^{k + 1}}\sum_{m = 0}^{n}\pars{1 + z}^{m} \,{\dd z \over 2\pi\ic} \\[3mm] & = \oint_{\verts{z} = 1}{1 \over z^{k + 1}}\, {\pars{1 + z}^{n + 1} - 1 \over \pars{1 + z} - 1}\,{\dd z \over 2\pi\ic} \\[3mm] & = \underbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{n + 1} \over z^{k + 2}} \,{\dd z \over 2\pi\ic}}_{\ds{n + 1 \choose k + 1}}\ -\ \underbrace{\oint_{\verts{z} = 1}{1 \over z^{k + 2}}\,{\dd z \over 2\pi\ic}} _{\ds{\delta_{k + 2,1}}} \\[8mm] \imp\ \color{#f00}{a_{n}} & = \fbox{\ds{\quad% {n + 1 \choose k + 1} - \delta_{k,-1}\quad}} \end{align}
\begin{align} \mbox{With}\ \pars{1}\,,\quad \color{#f00}{\sum_{m = 0}^{M}{m + k \choose k}} & = \bracks{{M + k + 1 \choose k + 1} - \delta_{k,-1}} - \bracks{{k \choose k + 1} - \delta_{k,-1}} \\[3mm] & = {M + k + 1 \choose k + 1} - {k \choose k + 1} \end{align} Thanks to $$\ds{@robjohn}$$ user who pointed out the following feature: $${k \choose k + 1} = {-k + k + 1 - 1 \choose k + 1}\pars{-1}^{k + 1} = -\pars{-1}^{k}{0 \choose k + 1} = \delta_{k,-1}$$ such that $$\begin{array}{|c|}\hline\mbox{}\\ \ds{\quad\color{#f00}{\sum_{m = 0}^{M}{m + k \choose k}} = \color{#f00}{{M + k + 1 \choose k + 1} - \delta_{k,-1}}\quad} \\ \mbox{}\\ \hline \end{array}\\$$
• Since $k=-1$ is covered in the first part, it should be noted that since $\binom{-1}{0}=1$, $$\binom{k}{k+1}-\delta_{k,-1}=0$$ therefore the final answer seems it should be $$\binom{M+k+1}{k+1}-\delta_{k,-1}$$ – robjohn Jul 25 '16 at 13:00
• @robjohn Thanks. I'm checking everything right now. – Felix Marin Jul 25 '16 at 21:48
• @robjohn Thanks. Fixed. – Felix Marin Jul 25 '16 at 22:09

This is essentially the same as the induction answer already mentioned, but it brings a pictorial perspective so I thought to add it as an alternative answer here.

Here's a restatement of the identity (which you can verify to be equivalent easily): On Pascal's triangle, start from a number (one of the $$1$$s) on the left edge and move diagonally rightwards and downwards, summing the terms as we go along. We can decide to stop at any point, and the sum of all these terms will be the number directly below and to the left of the final summand.

This actually trivialises the proof of the identity. Note that if we decided to add one more term to the sum (the term to the right of the current sum), on the one hand this "lengthens" the stick by $$1$$ tile, but on the other hand it adds the term adjacent to the sum---which by definition of Pascal's triangle, produces the number in the tile directly below and horizontally in between the sum and this new term. This can be rigorously translated to the inductive step in a formal induction proof.

To illustrate, let's refer to the picture in the question, and focus on the yellow hexagonal tiles. (Note that this is a reflected case of what I described above since it starts from the right edge, but this doesn't affect the discussion.) Currently, we have $$1+6+21+56=84$$, which is a true identity. If I added $$126$$ to the LHS, I would also be adding $$126$$ to the RHS, which by definition gives us $$210$$, the term below and in between them on Pascal's triangle. Once you really convince yourself of the validity of this argument, the (formal) proof of the identity should come naturally!

• Your answer emphasises the recursive nature of the solution. +1. But, you never seemed to hint towards the term of recursion in your answer! Answer at: math.stackexchange.com/a/4009354/424260 , states recursion, but using arithmetic. – jiten Feb 3 at 1:30