Binomial coefficients identity: $\sum i \binom{n-i}{k-1}=\binom{n+1}{k+1}$

Question

I am trying to prove

$ \sum_{i=1}^{n-k+1} i \binom{n-i}{k-1}=\binom{n+1}{k+1} $

Whichever numbers for $k,n$ I try, the terms equal, but when I try to use induction by n, I fail to prove the induction step:

Assume the equation holds for $n$. Now by Pascal's recursion formula,

$ \binom{n+2}{k+1}=\binom{n+1}{k+1} + \binom{n+1}{k}\\ =\sum_{i=1}^{n-k+1} i \binom{n-i}{k-1}+\binom{n+1}{k}, $

by induction assumption. In order to complete the proof, I would need to show

$ (n-k+2) \binom{n-(n-k+2)}{k-1} = \binom{n+1}{k} $

but the left-hand side is zero. What am I doing wrong?

EDIT:

There were links to similar questions when my question was marked as duplicate. However, these links are now gone, so I add them here as they were useful to me:

• How do i reduce this expression of binomial coefficients -- hint to Vandermont's identity
• Closed form for a formula with a summation over i(n−ik−1), and combinatorial proof? -- basically the same question, but two days earlier.

(I did search, but did not found these.)

Answer 1

Let's give a combinatorial proof.

$\binom{n+1}{k+1}$ is the number of subsets of $[n] = \{0,1,2, \ldots, n\}$ having exactly $k+1$ elements. Looking at the smallest or largest element of such a subset makes sense.

Let's look at the second smallest element instead. It can be any number from $1$ (for subsets which have $0$ and $1$ as their two smallest elements) to $n-k+1$ (for subsets which have $n-k+1, \ldots, n-1, n$ as their $k$ largest elements).

So $$\binom{n+1}{k+1} = \sum_{i=1}^{n-k+1} \textrm{number of subsets having $i$ as 2nd smallest element}.$$

So how many $(k+1)$-element subsets of $[n]$ have $i$ as 2nd smallest elements?

Well, the recipe to make such a subset is clear:

1. Chose any number in $0, 1, \ldots, i-1$ to serve as smallest element.
2. Pick $i$ as second smallest element.
3. Pick any $(k-1)$-element subset of $\{i+1, \ldots, n\}$ to serve as the $n-1$ elements larger than $i$.

In the first step, you have $i$ choices, the second step involves no choice and you have $\binom{n-i}{k-1}$ choices in the last step, because $\{i+1, \ldots, n\}$ has $n-i$ elements.

So the number of $(k+1)$-element subsets of $[n]$ having $i$ in second position is $i \binom{n-i}{k-1}$, which proves that $$\binom{n+1}{k+1} = \sum_{i=1}^{n-k+1} i \binom{n-i}{k-1}.$$

A slight generalisation of this proof shows that if $a+b=k$ $$ \binom{n+1}{k+1} = \sum_i \binom{i}{a}\binom{n-i}{b}:$$ you only have to look at the $(a+1)$-th largest element of the $(k+1)$-element subsets of $n+1$. We have just done the $a=1$ case.

Answer 2

No, to complete the proof along these lines you need to show that

$$\sum_{i=1}^{n-k+2}i\binom{n+1-i}{k-1}=\sum_{i=1}^{n-k+1}i\binom{n-i}{k-1}+\binom{n+1}k\;;\tag{1}$$

you forgot that the upper number in the binomial coefficient changes when you go from $n$ to $n+1$.

You can rewrite $(1)$ as

$$(n-k+2)\binom{k-1}{k-1}+\sum_{i=1}^{n-k+1}i\left(\binom{n+1-i}{k-1}-\binom{n-i}{k-1}\right)=\binom{n+1}k\;,$$

whose lefthand side reduces to

$$n-k+2+\sum_{i=1}^{n-k+1}i\binom{n-i}{k-2}$$

by Pascal’s identity. This in turn can be rewritten as

$$n-k+2+\sum_{i=1}^{n-k+2}i\binom{n-i}{k-2}-(n-k+2)\binom{k-2}{k-2}=\sum_{i=1}^{n-k+2}i\binom{n-i}{k-2}\;.$$

If you took the right induction hypothesis — namely, that the equation holds for $n$ and all $k$ — then your induction hypothesis allows you to reduce this last summation to a single binomial coefficient, which proves to be the one that you want.

Answer 3

The answers have been given already, but just for seeing a direct computation I post my answer:

$$\sum_{i=1}^{n-k+1}i\binom{n-i}{k-1}=\sum_{i=k-1}^{n}\left(n-i\right)\binom{i}{k-1}$$ $$=n\sum_{i=k-1}^{n}\binom{i}{k-1}-\sum_{i=k-1}^{n}\binom{i}{k-1}i$$

Now observe that:

$$\sum_{k=n}^{m}\binom{k}{n}k=\sum_{k=0}^{m}\binom{k}{n}k$$$$=\sum_{k=0}^{m}\binom{k-1}{n-1}k+\sum_{k=0}^{m}\binom{k-1}{n}k=n\sum_{\color{red}{k=0}}^{m}\binom{k}{n}+\left(n+1 \right)\sum_{\color{blue}{k=0}}^{m}\binom{k}{n+1}$$$$=n\sum_{\color{red}{k=n}}^{m}\binom{k}{n}+\left(n+1 \right)\sum_{\color{blue}{k=n+1}}^{m}\binom{k}{n+1}$$$$=n\binom{m+1}{n+1}+\left(n+1 \right)\binom{m+1}{n+2}\;\;\;\;\;\;\;\;\;\;$$

Using this we have:

$$=n\binom{n+1}{k}-\left[\left(k-1 \right)\binom{n+1}{k}+k \binom{n+1}{k+1}\right]$$ $$=n\binom{n+1}{k}-\left[k\left( \binom{n+1}{k}+\binom{n+1}{k+1}\right)-\binom{n+1}{k} \right]$$$$=n\binom{n+1}{k}-\left[k \binom{n+2}{k+1}-\binom{n+1}{k} \right]$$$$=\left(n-k\ \frac{n+2}{k+1}+1 \right)\color{red}{\binom{n+1}{k}}$$$$=\frac{n-k+1}{k+1}\color{red}{\binom{n+1}{n-k+1}}=\frac{n+1}{k+1}\color{red}{{\binom{n}{k}}}$$$$={\binom{n+1}{k+1}}$$

Hence we showed that:

$$\bbox[5px,border:2px solid #00A000]{\sum_{i=1}^{n-k+1}i\binom{n-i}{k-1}={\binom{n+1}{k+1}}}$$

Which is the claim.

Answer 4

A proof in terms of generating functions is quite direct, using binomial reflection identities (where $n\choose k$ is defined as $\frac{\Gamma(n+1)}{\Gamma(k+1)\cdot\Gamma(n-k+1)}$, accounting for the multiplicity of $0$).

The summation is rewritable as a convolution, $$\begin{aligned}&\sum_{i=1}^{n-k+1}\left(i\cdot{n-i\choose k-1}\right)\\ =&\sum_{i=0}^{n-k}\left((i+1)\cdot{k-1+(n-k-i)\choose k-1}\right)\\ =&\underset{\Sigma i=n-k}{\mathrm{conv}}\left({i+1\atop{k-1+i\choose k-1}}\right)\end{aligned}$$ Since ${k-1+i\choose k-1}=\frac{k}{i}\cdot{k-1+i\choose k}=(-1)^k\cdot\frac{k}{i}\cdot{-i\choose k}=(-1)^i\cdot{-k\choose i}$, we have $$\begin{aligned}=&\underset{\Sigma i=n-k}{\mathrm{conv}}\left({i+1\atop(-1)^i\cdot{-k\choose i}}\right)\\ =&\left[x^{n-k}\right]\left(\frac{1}{(1-x)^2}\cdot\frac{1}{(1-x)^k}\right)\\ =&(-1)^{n-k}\cdot{-2-k\choose n-k}\end{aligned}$$ and since in general ${-n\choose k}=(-1)^k\cdot{k-1+n\choose k}$, this equals ${n+1\choose n-k}$