I am looking for a closed form for the sum below  $$ \sum_{k=0}^{r} \frac{1}{2-\delta_{k, r-k}} \binom{n}{k} \binom{n}{r-k} $$  where $\delta_{k, r-k}$ is the Kronecker delta. I'd appreciate any helps in finding a closed-form solution for this sum.   To my understanding, this is basically equivalent to finding a closed-form solution to the following problem:  Given a set $S$ with size $|S|=n$, what is the number of two-element subsets of $\mathcal{P}(S)$, $\{A, B\} \in \mathcal{P}(S)$ such that $|A|+|B| = r$  for a fixed $r$.  I tried to use Wolfram Alpha, but it is given me this for the case where we count ordered pairs $(A, B)$ instead of sets (i.e. the sum if we ignore $\frac{1}{2-\delta_{k, r-k}}$), which contains factorial of a negative integer(!!) (and I don't know how to specify in Wolfram that $n&gt;r&gt;0$ and both are integers):  $$ \frac{(-1)^r (-2n+r-1)!}{(-2n-1)!r!} $$
7838
How can we prove that     $$\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}?$$   (Presumptive) Source: Theoretical Exercise 8, Ch 1, A First Course in Probability, 8th ed by Sheldon Ross.
51481
I used WolframAlpha to calculate a sum but it didn't show me the way :( Anybody has a hint or a solution for proving this sum? $$\sum_{n=0}^N\binom{2N-n}N2^n(n+1)=(1+2N)\binom{2N}N$$
12692
What is the sum of number of ways of choosing $n$ elements from $(n+r)$ elements where $r$ is fixed and $n$ varies from $1$ to $m$ ? Can this be reduced to a formula ?  $$ \sum ^m _{n=1} \binom{n + r}n $$
41291
How do I go about simplifying this:  $$\sum_{k=1}^nk \cdot k!$$  Wolfram alpha tells me it's the same as $(n+1)!-1$ but I don't see how.
10211
Could somebody explain to me where these two formulas come from as applications of the binomial theorem?  $$\sum_{k=0}^n {n \choose k}(-1)^kk^r=0$$ for non-negative integers $r\lt n$. And $$\sum_{k=0}^n {n \choose k}(-1)^kk^n=(-1)^nn!$$
103619
I am looking for a closed form (ideally expressed as elementary functions) of the function $\exp_n(x) = \sum_{k=0}^n x^k / k!$.  I am already aware of expressing it in terms of the gamma function. Background / Motivation When counting combinations of objects with generating functions, it is useful to be able to express the partial sum $1 + x + \cdots + x^n$ as $\frac{1-x^{n+1}}{1-x}$. For example, to count the number of ways to pick 5 marbles from a bag of blue, red, and green marbles where we pick at most 3 blue marbles and at most 2 red marbles, we can consider the generating function $f(x) = (1+x+x^2+x^3)(1+x+x^2)(1+x+x^2+\cdots)$. By using the partial sum identity, we can express it as $f(x) = \left(\frac{1-x^4}{1-x}\right)\left(\frac{1-x^3}{1-x}\right)\left(\frac{1}{1-x}\right)$.  Simplify, express as simpler product of series, and find the coefficient of the $x^5$ term. I want to be able to do the same for a generating function in the form $g(x) = \exp_{n_1}(x)^{p_1} \exp_{n_2}(x)^{p_2} \cdots \exp_{n_j}(x)^{p_j}$ The easiest way to extract the coefficient of a given term $x^p / p!$ would be to use a similar closed form expression for $\exp_n(x)$ and a similar technique to $f$. Attempted Solutions Differential equation Recall that the way to prove the identity $1+x+x^2+\cdots+x^n = \frac{1-x^{n+1}}{1-x}$ is to define $S = 1 + x + x^2 + \cdots + x^n$ and notice that: $S - Sx = 1 - x^{n+1}$. Likewise, notice that $y(x) = \exp_n(x)$ satisfies $y - y' = x^n/n!$.  Via SAGE, the solution is $y(x) = \frac{c+\Gamma(n+1,x)}{n!}e^x$.  Our initial condition $y(0) = 0$ so $c=0$.  By , $\Gamma(n+1,x) = n! e^{-x} \exp_n(x)$ so we have $y(x) = \exp_n(x)$. Recurrence Relation Notice that $\exp_n(x) = \exp_{n-1}(x) + x^n/n!$. Using the unilateral Z-Transform and related properties, we find that $\mathcal{Z}[\exp_n(x)] = (z e^{x/z})/(z-1)$. Therefore, $\exp_n(x) = \mathcal{Z}^{-1}\left[(z e^{x/z})/(z-1)\right] = \frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz$. $(z^n e^{x/z})/(z-1)$ has two singularities: $z = 1$ and $z = 0$.  The point $z = 1$ is a pole of order one with residue $e^x$.  To find the residue at $z = 0$ consider the product $z^n e^{x/z} (-1/(1-z)) = -z^n \left( \sum_{m=0}^\infty x^m z^{-m} / m! \right) \left( \sum_{j=0}^\infty x^j \right)$.  The coefficient of the $z^{-1}$ term is given when $n - m + j = -1$. The residue of the point $z=0$ is then $-\sum_{m,j} x^m / m! = -\sum_{m=n+1}^\infty x^m / m!$. Let $C$ by the positively oriented unit circle centered at the origin.  By Cauchy's Residue Theorem, $\frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz = \frac{1}{2 \pi i} 2 \pi i \left(e^x - \sum_{m=n+1}^\infty x^m / m!\right) = \exp_n(x)$. Finite Calculus I've tried to evaluate the sum using finite calculus, but can't seem to make much progress.
114867
How to get a closed form for $$\sum_{k=0}^{n-p}\binom{n}{k}\binom{n}{p+k}\;?$$  I tried to write binominal in term of gamma function but I got no result. What is your suggestion to solve the problem?
27799
Can someone tell me how to get a closed form for  $$\sum_{k=1}^n k^p$$  For $p = 1$, it's just the classic $\frac{n(n+1)}2$.   What is it for $p &gt; 1$?
99071
I typed into my Mathematica:$\sum _{n=1}^{\infty } \frac{1}{n^2+1}$ , and the result was: $$\frac{-1+\pi \coth (\pi)}{2}$$ I know how to estimate the aforementioned sum , but I have no idea how to get its closed form.
54720
Mathematica tells me that  $$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}.$$  Although I have not been able to come up with a proof.  Proofs, hints, or references are all welcome.
32926
I am trying to find a closed form solution for $\sum_{n\ge0} nx^n\text{, where }\lvert x \rvert&lt;1$.  This solution makes sense to me:  $\sum_{n\ge0} x^n=(1-x)^{-1} \\ \frac{d}{d x} \sum_{n\ge0} x^n = \frac{d}{d x} (1-x)^{-1} \\ \sum_{n\ge0} nx^{n-1} = (1-x)^{-2} \\ x \sum_{n\ge0} n x^{n-1} = x(1-x)^{-2} \\ \sum_{n\ge0} nx^n=\frac x{(1-x)^2}$  However, a  I am reading used the following method:  $$\sum_{n\ge0}nx^n=\sum_{n\ge0}x\frac d{dx}x^n= x\frac d{dx}\sum\limits_{n\ge0}x^n=x\frac d{dx}\frac1{1-x}=\frac x{(1-x)^2}$$  This seems closely related to the solution I described above, but I am having difficulty understanding it.  Can someone explain the method being used here?
28357
While I've studied integrals involving  I've observed that  $$\int_0^1 \operatorname{Li}_p(x) \, dx \stackrel{?}{=} \sum_{k=2}^p(-1)^{p+k}\zeta(k)+(-1)^{p+1},\tag{1}$$  for any integer $p\geq2$. Here $\zeta$ is the .  After that I have three questions.   $1^\text{st}$ Question. Is $(1)$ true? If it is, how could we prove it? $2^\text{nd}$ Question. If it's a well-known result could you give any reference? $3^\text{rd}$ Question. I think there is also a similar closed-form of $\int_0^b \operatorname{Li}_p(x) \, dx$, for any integer $b \geq 1$. What is the closed-form of this integral?
47143
EDIT: Really sorry to put you through the trouble. The expression I am looking for is actually the summation of these values. Not the product. My bad  I'm looking for a condensed form of $\sum_{i=0}^{N/2}{N-i\choose i}$.  E.g. for $n=6$, I want to reduce ${6\choose0}+{5\choose1}+{4\choose2}+{3\choose3}$ into a single NCR or something that can be computed easily. Is it possible?
20633
I found the solution of series on Wolfram Alpha    $ \sum\limits_{k=1}^{n} \left(\frac{1}{2k+1} - \frac{1}{2k+2}\right) = \sum\limits_{k=1}^{n} \frac{1}{(2k+1)(2k+2)} = \frac{1}{2} \left(H_{n+\frac{1}{2}} - H_{n+1} -1 + \text{ln}(4)\right)$  Can someone tell how to prove this in the form of Harmonic numbers?
61442
The binomial expression given as follows: $$\sum_{k=0}^{2n}\left(-1\right)^{k}\binom{2n}{k}^{2}$$ results nicely into the following closed form: $$(-1)^{n}\binom{2n}{n}$$ I wish to know how exactly is it being done? I haven't been able to make much progress in solving it. My approach: \begin{align} \sum_{k=0}^{2n}(-1)^{k}\binom{2n}{k}^{2} = \binom{2n}{0}^{2} - \binom{2n}{1}^{2} + \binom{2n}{2}^{2} - ... -\binom{2n}{2n-1}^{2} + \binom{2n}{2n}^{2} \\ = \binom{2n}{0}.\binom{2n}{0} - \binom{2n}{1}.\binom{2n}{1} + \binom{2n}{2}.\binom{2n}{2} - ... -\binom{2n}{2n-1}.\binom{2n}{2n-1} + \binom{2n}{2n}.\binom{2n}{2n} \\ \text{By Symmetry of binomial coefficients} \\ = \binom{2n}{2n}.\binom{2n}{0} - \binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n-2}.\binom{2n}{2} - ... -\binom{2n}{1}.\binom{2n}{2n-1} + \binom{2n}{0}.\binom{2n}{2n} \\ = \binom{2n}{2n}.\binom{2n}{0} + \binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n-2}.\binom{2n}{2} + ... +\binom{2n}{1}.\binom{2n}{2n-1} + \binom{2n}{0}.\binom{2n}{2n} - 2.\left(\binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n -3}.\binom{2n}{3} + ... + \binom{2n}{1}.\binom{2n}{2n-1}\right) \\ \text{By Vandermond's identity, first component, i.e.not enclosed within -2.(...) evaluates to C(4n, 2n)} \\ \binom{4n}{2n} - 2.(\binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n -3}.\binom{2n}{3} + ... + \binom{2n}{1}.\binom{2n}{2n-1}) \end{align} I'm lost beyond this point. It will be extremely helpful if someone can direct me in the right direction or provide the answer to this perplexing and challenging problem. Thank you.
103041
Consider this infinite sum of matrices. Is there any closed form to express this sum?   $S=B+ABA^T +A^2B({A^T})^2+A^3B({A^T})^3+...$  And B is diagonal.  Thanks
68610
What is the closed form of  $$\sum_{k=0}^n \frac{x^k}{k!}$$  as a function of $x$ and $n$?  Knowing that it converges to $e^x$ when $n\to \infty$.