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


*

*Chose any number in $0, 1, \ldots, i-1$ to serve as smallest element.

*Pick $i$ as second smallest element.

*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.
A: 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.
A: 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.
