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Here's another proof. We first rewrite the identity (by setting $k_i=j_i+1$) as $$ \sum_{j_1+\cdots +j_n=K-n}\prod_{i=1}^n \frac{(j_i+1)^{j_i-1}}{j_i!} = n\frac{K^{K-n-1}}{(K-n)!}. \tag{1} $$ Let …

The second sum is equal to $\binom{n-1}{n-k}$. For each fixed $a$ the first sum can be evaluated but I don't think there's a nice general formula. For example, for $a=1$, Maple gives $$-\frac{{\binom{ …

$\def\des{\operatorname{des}}$Let $\des(\pi)$ be the number of descents of the permutation $\pi$. Then for any permutation $\pi$ in $S_k$, we have \begin{equation*}\binom{xy+k-\des(\pi)-1}{k} =\sum_{\ …

The formula (which holds for $t>1$ but not for $t=1$), is equivalent to $$\sum_{t=1}^\infty C_{t-1}\bigl(x(1-x)\bigr)^t = x,$$ which follows directly from the generating function $$\sum_{t=1}^\infty C …

It is not hard to show that the factor $n+1$ is present. Theorem. Let $p(i)$ be a polynomial. Then there exists a polynomial $P(n)$ such that for $i\ge0$ we have $\sum_{i=0}^n p(i)= P(n)$, and $P(n)$ …

A closed form is \begin{equation*} \sum_{i=j+1}^{n} \binom{\binom{i}{j}}{2} =\sum_{k=1}^j \frac{1}{2}\binom{j}{k}\binom{j+k}{k}\binom{n+1}{j+k+1}.\tag{1}\label{474985_1} \end{equation*} For fixed $j$ …

Here's a straightforward proof, using generating functions, though it's not as elegant as Ofir's. We have $$ \sum_{a=0}^\infty\binom{a}{j}x^a =\frac{x^j}{(1-x)^{j+1}}. $$ Setting $j=(p-1)k$ and summin …

The first formula is a special case of one of the standard hypergeometric series summation formulas called Kummer's theorem. (See, e.g., http://mathworld.wolfram.com/KummersTheorem.html or http://en …

The two identities are both special cases of Vandermonde's theorem (also called the Chu-Vandermonde theorem), which is the most well-known binomial coefficient identity after the binomial theorem. The …

In terms of hypergeometric series, the sum is $_3F_2(-n, 1+n, 1/2;1,3/2;1)$ and the identity is a special case of Saalschütz's theorem (also called the Pfaff-Saalschütz theorem), one of the standard h …

There's nothing special about the gamma function; the failure of the limit to exist when $a$, $b$, and $n$ are negative is exactly the same as the failure of $$\lim_{(x,y,z)\to(0,0,0)} \frac{xy}{z}$$ …

Just as the Fubini numbers are the coefficients of the exponential generating function $$\sum_{k=0}^\infty (e^x-1)^k = \frac{1}{2-e^x},$$ the exponential generating function for the alternating sum is …

Here is a combinatorial proof. It is more convenient to prove the equivalent formula (obtained by setting $n=m+2j$) $$\sum_k \binom{k}{j}\binom{m+2j}{2k}=2^{m-1}\frac{m+2j}{m+j}\binom{m+j}{j}.\tag{$*$ …

Although this isn't an answer to the question, it's worth pointing out that the second and third families are essentially binomial coefficients. We have $$U_2(m,n):=\frac{(2m)!\,(2n)!}{m!\, n!\, (m+n) …

More generally, $$\sum_{k\ge1} \frac{k}{m}\binom{2m}{m-k}\cdot\frac{k}{n} \binom{2n}{n-k} = C_{m+n-1}.$$ This can be proved by the same reasoning as in Timothy Budd's answer. This formula gives the LD …