Proof of the hockey stick/Zhu Shijie identity $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$

Question

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

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?

Thanks for your help.

---

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.

Answer 1

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

Answer 2

You can use induction on $n$, observing that

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

Answer 3

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:

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

Answer 4

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

Answer 5

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.

Answer 6

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.

Answer 7

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.)

Answer 8

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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 $$ \begin{array}{|c|}\hline\mbox{}\\ \ds{\quad\color{#f00}{\sum_{m = 0}^{M}{m + k \choose k}} = \color{#f00}{{M + k + 1 \choose k + 1} - \delta_{k,-1}}\quad} \\ \mbox{}\\ \hline \end{array}\\ $$

Answer 9

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 using $(1)$ again we have $$\sum_{t=0}^{n}\dbinom{t}{k}=\dbinom{n+1}{k+1}-0=\color{red}{\dbinom{n+1}{k+1}.}$$

Answer 10

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.

Answer 11

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.

Answer 12

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

Answer 13

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.

Answer 14

Let me attempt a “story” solution:

Let’s say you are trying to find ways to select $k+1$ integers from the first $n+1$ integers.

We consider all the cases, from where the largest number is $k+1$, up to where the largest number is $n+1$.

It is evident that if the largest number is $k+1$, that there is only one possibility, which is selecting every number from $1$ to $k+1$, and we can write this as $\binom kk$.

More generally, if the largest number is $j\geq k+1$, there are $\binom{j-1}{k}$ selections.

We then sum all of these up, all the way to where the largest number is $k+1$ itself.

Answer 15

First proof

Using stars and bars, the number of ways to put $n$ identical objects to $k$ bins(empty bin allowed) is $\binom{n+k-1}{k-1}$.

If we reduce the number of bins by one, the number of ways to put $n$ identical objects to $k-1$ bins is $\binom{n+k-2}{k-2}$.

We can also count the number of ways to put $n$ identical objects to $k$ bins using the answer for $k-1$ bins. Split them into different cases based on how many objects you put in the first bin. \begin{array} {|r|r|}\hline \text{Number of objects for first bin} & \text{Number of objects remaining for other bins} & \text{Number of ways to distribute} \\ \hline 0 & n & \binom{n+k-2}{k-2} \\ \hline 1 & n-1 & \binom{n+k-3}{k-2} \\ \hline 2 & n-2 & \binom{n+k-4}{k-2} \\ \hline \vdots & \vdots & \vdots \\ \hline n-2 & 2 & \binom{k}{k-2} \\ \hline n-1 & 1 & \binom{k-1}{k-2} \\ \hline n & 0 & \binom{k-2}{k-2} \\ \hline \end{array}

Therefore, $$\binom{n+k-1}{k-1} = \binom{k-2}{k-2} + \binom{k-1}{k-2} + \binom{k}{k-2} +\dots+ \binom{n+k-4}{k-2} + \binom{n+k-3}{k-2} + \binom{n+k-2}{k-2}$$

Rename the variables: Let $m+1 = n+k-1$ and $r+1 = k-1$. We get: $$\binom{m+1}{r+1} = \binom{r}{r} + \binom{r+1}{r} + \binom{r+2}{r} +\dots+ \binom{m-2}{r} + \binom{m-1}{r} + \binom{m}{r}$$

This is the identity when expanded.

Second proof

Suppose we have $m+1$ people where everyone has different age and we want to select $r+1$ people to be in some committee. There are $$\binom{m+1}{r+1} $$ ways to form the committee.

Alternatively, we can count this by splitting the committee based on who the oldest person in the committee is.

We can have the oldest person to be the oldest in the committee. Then there are $m$ people left and we still need to add $r$ people to be in the committee. So there are $\binom{m}{r}$ ways.

We can have the second oldest person to be the oldest in the committee. Then there are $m-1$ people left since we must exclude the first oldest person and second oldest(already selected) and we still need to add $r$ people to be in the committee. So there are $\binom{m-1}{r}$ ways.

In general, suppose we have the kth oldest person to be the oldest in the committee. Then there are $m-k+1$ possible people left to add(excluding kth oldest) since we must exclude the oldest members up to the $k$th oldest person. We still need to add $r$ people to be in the committee. So there are ${m-k+1 \choose r}$ ways.

In each case, there needs to be at least $r$ people left to choose so the last case is when we select the $(m+1-r)$th oldest person. This will give us ${r \choose r}$ ways.

So $$\binom{m+1}{r+1} = \binom{r}{r} + \binom{r+1}{r} + \binom{r+2}{r} +\dots+ \binom{m-2}{r} + \binom{m-1}{r} + \binom{m}{r}$$