I cant wrap my head around simplifying the following sum:     $$\sum_{k=0}^n \binom{n}{k}x^k(1-x)^{n-k}k,$$   where $0&lt;x&lt;1$.    I tried to apply standard formulas .
Why does:   $$ \sum_{k=0}^{n} k \binom nk p^k (1-p)^{n-k} = np $$ ? Taking the derivative of:   $$ \sum_{k=0}^{n}  \binom nk p^k (1-p)^{n-k} = (1 + [1-P])^n = 1 $$   does not seem useful, since you would get zero. And induction hasn't yet worked for me, since – during the inductive step – I am unable to prove that:   $$ \sum_{k=0}^{n+1} k \binom {n+1}{k} p^k (1-p)^{n+1-k} = (n+1)p  $$   assuming that:   $$ \sum_{k=0}^{n} k \binom {n}{k} p^k (1-p)^{n-k} = np  $$
Dropdownlist control with s for asp.net (webforms)? Can anyone recommend a dropdownlist control for asp.net (3.5) that can render option groups? Thanks Nested Mathematical Functions This question has bugged me for a while. I want to write a nested function, f(g(x)). If I naively enter it:  \begin{align}     f(g(x))=y \end{align}     The problem is that each and every parenthesis is the same size as all of the others. Shouldn't the outer parentheses be larger? But if I do this:  \begin{align}     f\left(g(x)\right)=y \end{align}     Now it looks even less like the functions are nested. What's the proper way to do this? Semicolons at the end of list items In school, we were taught to end each list item with a semicolon. But I have never actually seen this. Can somebody please confirm this? Should we end each list item with anything at all?  I've seen people use periods after each list item, colons, and even commas. But I've never seen a semicolon at the end of each item — and even if I have, it would be that rare that I've forgotten.  Example:   item; item; item; item From 2012 onward the SQL Server docs show they support  which I'm trying to use instead of a LIMIT.  The following works fine in PostgreSQL to sample a result set,  SELECT * FROM ( VALUES (1),(2),(3) ) AS t(x) OFFSET 0 ROWS FETCH NEXT 1 ROWS ONLY;   However, with SQL Server, I get  Msg 153, Level 15, State 2, Line 4 Invalid usage of the option FIRST in the FETCH statement.   What's going on here? Does SQL Server support the standardized OFFSET .. FETCH? $$\int_{-\infty}^{\infty} \frac{\cos(αx)}{(x^2+1)(x^2+4)} \mathrm dx. $$  I am not sure how to solve this question. Can anyone help me to approach this problem. Thanks. How are NFL matchups determined for each season? During the 2012 NFL Season, the Pittsburgh Steelers and Baltimore Ravens play each other in weeks 11 and 13, two weeks (in both NFL schedule and time) apart. To me, that seems too close for comfort.   I do know that teams play other teams in their division twice. How does the NFL schedule procedure work? Overleaf does not compile after using Biblatex I have my PhD deadline on Monday and disaster has struck at the very last minute! Any assistance would be eternally appreciated.  I'm using Overleaf (v2) to write my thesis - it's a very large file tree containing multiple .tex files.  Everything was compiling fine, until I decided to experiment and use:  \usepackage{biblatex}   It compiled with this included but looked odd, so I deleted the line again. However, now it does not compile - despite the document being exactly the same as before the problem began. Existence of elements of order $p^k$ in a finite group By Cauchy's theorem, if a prime number $p$ divides the order of a group $G$, then there is an element $g$ in $G$ whose order is $p$. In addition, if $|G|$ admits a prime decomposition with a factor $p^k$, then Sylow's theorem tells us that a subgroup of order $p^k$ must exist.   My question is the following. If $|G|$ admits a prime decomposition with a factor $p^k$, does this imply that an element of order $p^k$ must exist in $G$?   Update: As explained in Mindlack's answer (among others), the implication does not hold and simple counterexamples exist. In the comments, Mindlack discusses special cases where such an element exists. Permanently removing apache2 It seems a simple apt-get remove apache2 does not completely remove apache2 as I can still see it on one of the processes when running top. How does one remove apache2 completely on his ubuntu server?  It's not removed indeed:  ~# which apache2 /usr/sbin/apache2 ~# whereis apache2 apache2: /usr/sbin/apache2 /etc/apache2 /usr/lib/apache2 /usr/share/apache2 /usr/share/man/man8/apache2.8.gz   But when I do apt-get remove apache2 again:  # apt-get remove apache2 Reading package lists... Done Building dependency tree Reading state information... Done Package apache2 is not installed, so not removed 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded. Plot electric potential and field I am a precision modeler, which means I like to be exact instead of "eyeballing" things when I model. I am very happy to see the option to Edge Slide with Clamping Off in v2.75a, but I don't seem to be able to snap the edge slide destination to a vertex point elsewhere on the model. Is this possible and I am just missing something? What is the coefficient of the term $a^i b^j c^k$ in the expansion of $(a+b+c)^n$? What is the coefficient of the term $a^i b^j c^k$  in the expansion of   $(a+b+c)^n$ ?   Given solution:     Before collecting like terms, each term in the expansion is a product   of $n$ factors each  of which is $a, b,$ or $c$.  Those terms which   have exactly i a’s, j b’s and k c’s can be combined into   the single term $a^i b^j c^k$ and the number of such terms will be   equal to the number of n letter strings containing i a’s, j b’s and k   c’s, which by the Mississippi rule is $\frac{n!}{(i!\times j! \times  k!)}$.   I understand that this problem is actually about the derivation of the coefficient of the multinomial theorem,but I am not sure about the combinatoric reasoning,what I mean is that I understand the analogy problem but what I couldn't is how the analogy holds here,Could anybody explain this to me? I asked a  on stackoverflow at the end of last year for a beta or pre-release product and got the answer I wanted.  3 months later someone else answered the question with the correct answer for a newer release.  This answer got way more upvotes than the one I accepted.  I only noticed this today so changed the accepted answer.  Can I get a notification email when an answer gets say 10 more votes than the one I accepted.  Just to let me know something might be up. (This question is only as reference, I will add the answer myself)  Suddenly the search in the expl3 documentation interface3.pdf no longer works. As an example searching for "divis" gives this as the first hit:    In sumatra there is no problem, other files (e.g. source3.pdf) works fine. What could be the reason?) show that Cov(X+ Y, X-Y)= Var(X) - Var(Y) Let's have 2 numbers, N and K, where K divides N. The number of K-combinations from a given set S of N elements is a well known formula. Let's concatenate N/K groups (resulting in N elements) such as the resulting set is N. How many possibilities are there, i.e. what's the formula?  For instance:    N=4, K=2, C(4, 2) = 6 {1,2},{1,3},{1,4},{2,3},{2,4},{3,4}   The 3 possibilities are:     {1,2},{3,4}   {1,3},{2,4}   {1,4},{2,3}   I generated these combinations and I think the number goes like this:  (4, 2): 3 (6, 3): 10 (6, 2): 15 (10, 5): 126 (9, 3): 280 (10, 2): 945 (14, 7): 1716 (12, 4): 5775 (15, 5): 126126 (15, 3): 1401400   Apparently, the result always divides with (N-1). Modify footer used by elsarticle.cls When is the Killing form null? When is the Killing form $\kappa$ of a Lie algebra $\mathfrak g$ null, i.e. $\kappa(\cdot,\cdot)=0$? Surely this is true for any Lie algebra with trivial bracket, but is this the only case? I can't seem to find any nontrivial examples. If $\sin \theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$, then $\sin 2\theta$ is a root of $x^2 -44x +36=0$ My own bonafide attempt. $$ 0&lt;\theta&lt;\pi/2$$   and $$\sin\theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$$   then show that $\sin 2\theta$ is a root of the equation $$x^2 -44x +36=0$$   I tried to use the above given equation of all the trigonometric ratios, though I ended up with an expression of $\sin\theta\cos\theta$.   But this too is in the form of $\sin\theta$ and $\cos\theta$ which was $\sin\theta\cos\theta=\frac{1}{6-\sin\theta-\cos\theta}$.  When I put that in the quadratic equation, it would again transform into the form of $\sin\theta$ and $\cos\theta$, and hence at last I couldn't prove the thing.  Even hints would work, as I would like to solve the question myself. "Leave a comment" link even when you can't On  I disabled  comments for "USDCAD options vs CADUSD options arbitrage?", but it  still has a "Leave a Comment" link.   Clicking the link takes you to    where it, of course, shows that you can't leave a comment.   The timeline page correctly does not show the 'comment' link:    How to fix?