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I'm looking for an algebraic proof of this identity for $n, k \in \mathbb{N}$:

$$\sum\limits_{i=0}^n \binom{i}{k} = \binom{n + 1}{k + 1}$$

So far, I've turned the left hand side of the equality into $\frac{1}{k!}\sum\limits_{m=0}^{n-k} \frac{(m+k)!}{m!}$, but I got stuck there and can't get any further.

Martin Sleziak
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APerson
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3 Answers3

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Let's examine $\displaystyle\sum_{k=0}^\infty\sum_{i=0}^n\binom{i}{k}x^k$. At the end, we'll look at the coefficient of $x^k$ to get our answer.

\begin{align} \sum_{k=0}^\infty\sum_{i=0}^n\binom ikx^k&=\sum_{i=0}^n\sum_{k=0}^\infty\binom ikx^k\\ &=\sum_{i=0}^n(1+x)^i\\ &=\frac{(1+x)^{n+1}-1}{(1+x)-1}\\ &=\frac{(1+x)^{n+1}-1}{x}\\ &=\frac{\sum_{j=0}^\infty\binom{n+1}j x^j-1}x\\ &=\frac{\sum_{j=\color{Red} 1}^\infty\binom{n+1}j x^j}x\\ &=\sum_{j=1}^\infty\binom{n+1}j x^{j-1}\\ &=\sum_{k=0}^\infty\binom{n+1}{k+1}x^k\\ \sum_{i=0}^n\binom{i}{k}&=\binom{n+1}{k+1} \end{align} The first step is The Most Powerful Proof Technique in Mathematics: switching the order of summation. (Which is why I had to put the extra $\sum$ sign in there. I can't switch the order of summation without two summations there in the first place.)

The $x^k$ term is there so I can get back the answer at the end. If I hadn't put it in there, I would have just ended up with $2^{n+1}-1$ at the end, which isn't very helpful.

Akiva Weinberger
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  • That technique works for a lot of sums, by the way. If there's another variable (i.e. $k$), multiply by $x^k$ and sum over it, and then switch the sigmas. If there isn't another variable (i.e. if the problem was $\sum\binom i{17}$), find a constant (i.e. $17$) and replace it by a variable, and then use the same strategy. – Akiva Weinberger Jul 27 '15 at 02:30
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    I wouldn't dare say this is the most powerful technique in mathematics. It is useful, though. – Pedro Aug 01 '15 at 22:45
  • @PedroTamaroff Yeah, I know. (I was quoting someone I know.) – Akiva Weinberger Aug 02 '15 at 07:56
  • This is just using generating functions, which is not necessary for an algebraic proof. In order to understand the identity, the OP can read the combinatorial proof. Personally, I think the answer below is more accessible and more informative. –  Mar 17 '20 at 04:06
  • I've seen this technique used a lot when using the Snake Oil Method, as seen in Wilf's "Generatingfunctionology" – Joshua Siktar Jun 22 '21 at 21:25
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Use induction on $n$. The relation $\displaystyle\sum_{i=k}^n\binom ik=\binom{n+1}{k+1}$ is trivially verified if $n=1$ (and $k\le n$).

Suppose, by inductive hypothesis, the formula is true for some $n\ge 1$ and $k\le n$. Then $$ \sum_{i=k}^{n+1}\binom ik=\sum_{i=k}^n\binom ik+\binom {n+1}k=\binom{n+1}{k+1}+\binom {n+1}k = \binom{n+2}{k+1}. $$ If $k=n+1$, then the left-hand side is reduced to just one term: $\dbinom{n+1}{n+1}$, which is is equal to $\dbinom{n+2}{n+2}$.

Bernard
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{i = 0}^{n}{i \choose k} & = \sum_{i = 0}^{n}\bracks{z^{k}}\pars{1 + z}^{\, i} = \bracks{z^{k}}\sum_{i = 0}^{n}\pars{1 + z}^{\, i} = \bracks{z^{k}}{\pars{1 + z}^{n + 1} - 1 \over \pars{z + 1} - 1} \\[5mm] & = \bracks{z^{k + 1}}\pars{1 + z}^{n + 1} - \bracks{z^{k + 1}}1 = \bbx{\large{n + 1 \choose k + 1} - \delta_{k,-1}} \\ & \end{align}

Felix Marin
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