I was given this identity for $n &gt; 0$, $$\sum_{k=0}^{n} (-1)^k {n \choose k} = 0$$  I need to explain why, using the "weirdo" method, there are exactly many subsets with an even number of elements as there are subsets with an odd number of elements, and then give a combinatorial proof of the identity.  I am stuck on getting started with the "weirdo" method. Can anyone get the ball rolling with this method's use?
I came across the following problem in a book:  Give a combinatorial proof of  $$ {n \choose 0} + {n \choose 2} + {n \choose 4} + \, \, ... \, = {n \choose 1} + {n \choose 3} + {n \choose 5} + \, \, ...$$ using the "weirdo" method (i.e., where one of the elements is chosen as special and included-excluded -- I'm sure you get the idea).  After days of repeated effort, the proof has failed to strike me. Because every time one of the elements is excluded, the term would be ${n-1 \choose k}$ and not $ {n \choose k}$, which is not the case in either of the sides of the equation.
Is it possible to copy an object in Illustrator so that if you make a change in any of them the changes will apply in both? Actually I have this situation where I have a lot of instances of the same object and I need to be faster on changes and copy-ing again and again is not a proper solution, but consumes a lot fo time.  Thank you for your time! Once a lead is converted it automatically creates an account, contact and one opportunity. But in my case i want to display the result without creating an Account. Please provide me with a sample code.  Any help will be appreciated... "does not contain a static 'main' method suitable for an entry point" Is it possible to send an email from my Java application using a GMail account?  I have configured my company mail server with Java app to send email, but that's not going to cut it when I distribute the application. Answers with any of using Hotmail, Yahoo or GMail are acceptable. Where does the idiom "beating around the bush" come from? Where does the idiom "beating around the bush" come from? Trim string in JavaScript? How do I trim a string in JavaScript? That is, how do I remove all whitespace from the beginning and the end of the string in JavaScript? What is JavaScript's highest integer value that a number can go to without losing precision? How to find files between two dates using "find"? This is a  about DNS/Hostnames resolution to IPs/Ports   Example 1  I'm running a web server on port 80 and another on port 87. I would like to use DNS so that www.example.com goes to port 87. How can I accomplish this using DNS only?  Example 2  I'm running a service on my server on a non-standard port. How can I get clients to connect to this non-standard port automatically? Can I use DNS? Is there some application specific support where DNS could indicate the IP and Port?   Example 3  Do some application protocols specifically support hostname awareness, and allow special actions to be taken based on this information? Are there other questions on Server Fault that cover some of these?  Commandeering: This question was originally asking about running IIS and Apache on the same server, but the same concepts can be applied to any server software receiving connections from clients. The Answers below describe the technical problems and solutions of using DNS and application protocol support to assign a port number for a client to connect. The question is to calculate the radius of convergence of $\sum\limits_{n=0}^\infty n!x^{n^2}$.  I dont know how to calculate R.O.C for such type of questions involving $n^2$ as power of x. How do we do this? Any help appreciated. Let $x = (a, b), y = (c, d) \in \mathbb{Z}^2$. What is the condition on $a, b, c, d$ so that ${x, y}$ is a basis?   My answer: $ad\neq bc$ and $gcd(a, c) = gcd(b, d) = 1$.  The first condition ensures that they aren't the same vector; the second ensures that we can actually "get" all of the integer values/lattice points.  Is this correct?  Thanks. I have a website that sells music artists posters. Each artists has a page containing a small bio, and a list of posters we are selling representing the artist.  Currently our HTML looks like that  &lt;h1&gt;{Artist name}&lt;/h1&gt; bio &lt;h2&gt;Posters&lt;/h2&gt; &lt;ul&gt;   &lt;li class="poster"&gt;...&lt;/li&gt;   ... &lt;/ul&gt;   We are redesigning this page, and we have a new template that we really like. Problem is that in this new version the &lt;h2&gt;Posters&lt;/h2&gt; does not appear anywhere on the page.  As you might expect, a significant part of our traffic comes from people typing something like : "{Artist name} poster" on Google.  I fear that, by removing the word "poster" from the page, we will reduce our chances of getting these people to come to the site, since Google might down rank us.  I see several options here :   Give Google a slightly altered page that contains the H2 tags. But I think this falls into the "Black Hat SEO" category. And I don't want that. Put the heading. Hide it with CSS. Everyone wins, including accessibility tools which can determine the structure of the page using the "invisible" heading. But I do not know if this is considered a good practice. Forget about the heading, and find other ways to tell Google that we really are about posters. (via keywords, but do they really have the same "weight" as an H2?) Adapt the design to our SEO need, and find a way to put that heading nicely.   Are there webmasters out there that have been facing the same issue, how did you resolve it? Some webpages forces mobile version even I change user-agent. I changed user-agent in built-in browser and "Dolphin Browser HD",  I am checking it on page www.user-agent-string.info  Main problem is engadget.com, always in mobile version. I tried to clean cookies, restart browsers etc.  This is my results:   in built-in browser, user-agent - "desktop"   () doesn't work, redirects to mobile version   in "Dolphin Browser HD", user-agent - "desktop"    (NSFW: redtube.com) doesn't work, redirects to mobile version  (allegro.pl) works fine  in "Dolphin Browser HD", user-agent - "IPad"   () doesn't work, redirects to mobile version  (allegro.pl) works fine (NSFW: redtube.com) works fine  in "Dolphin Browser HD", user-agent - "custom": Mozilla/5.0 (Windows NT 6.0; rv:7.0.1) Gecko/20100101 Firefox/7.0.1   (engadget.com]) doesn't work, redirects to mobile version  (NSFW: redtube.com) doesn't work, redirects to mobile version  (allegro.pl) works fine    Maybe this pages using "adobe flash" user-agent detection ?  (I'am using Asus Transformer TF101, Android 3.2.1, Adobe Flash Player 11) How to find all the used IP addresses on a network In my work there is a lot of computer and I want make a joke. I can close computer over the network but finding IP addresses is hard to me.  How can I easily find all online IP adresses from 192.168.1.aa to 192.168.1.zz? Query is slow in SQL Server 2014, fast in SQL Server 2012 During the migration of one of our databases from SQL Server 2012 (SP1, CU2) to SQL Server 2014 (SP1), we have experienced some strange issues.  One of the queries that completes within seconds on SQL Server 2012 seem to be hanging on SQL Server 2014.  SELECT  DISTINCT      src.[Id] FROM     [stg].[BaseVolumes] src JOIN          [tmp].[Dates] d ON src.[CalWeek_Nmbr] = d.[CalYrWkDense_Nmbr] WHERE        EXISTS (SELECT  *             FROM                     (SELECT     ctry.[ISOCode]                  AS  [Mkt_Code]                             , so.[Code]                     AS  [SlsOrg_Code_AK]                             , so.[DistributionChannelCode]  AS  [DistChnl_Code_AK]                             , prd.[SupplierCode]                AS  [SKU_Code_AK]                             , cl6.[Code]                        AS  [CstHierLvl06_Code_AK]                             , lp.[BaseDateID]                   AS  [Dte_EK]                  FROM   [PM_APP].[edw].[BaseVolumeDayCurrent]   lp                  JOIN   [PM_APP].[dbo].[Country]                ctry    ON  lp.[CountryID] = ctry.[ID]                  JOIN   [PM_APP].[dbo].[SalesOrganisation]      so      ON  lp.[SalesOrganisationID] = so.[ID]                  JOIN   [PM_APP].[dbo].[Product]                prd     ON  lp.[ProductID] = prd.[ID]                     JOIN   [PM_APP].[dbo].[CustomerLevel6]         cl6     ON  lp.[CustomerID] = cl6.[ID]                  WHERE                        lp.[ModifiedByApp] = 1) lkp             WHERE                    src.[Mkt_Code]                  =   lkp.[Mkt_Code]                 AND src.[SlsOrg_Code_AK]        =   lkp.[SlsOrg_Code_AK]                 AND src.[DistChnl_Code_AK]      =   lkp.[DistChnl_Code_AK]                 AND src.[SKU_Code_AK]           =   lkp.[SKU_Code_AK]                 AND src.[CstHierLvl06_Code_AK]  =   lkp.[CstHierLvl06_Code_AK]                 AND d.[Dte_EK]                  =   lkp.[Dte_EK]         )   Row count:   BaseVolumes:            23108  Dates:                  18628  BaseVolumeDayCurrent:   108115503  Country:                249 SalesOrganisation:      29 Product:                18446 CustomerLevel6:         295   Wait stats showing high CXPacket, but there is one hanging task with SOS_SCHEDULER_YIELD, (print from sys.sysprocesses)   spid | blocked | waittime | lastwaittype          | cpu     | physical_io   76   | 0       | 9886044  | CXPACKET              | 13902   | 31192  76   | 0       | 0        | SOS_SCHEDULER_YIELD   | 9829719 | 83077  76   | 0       | 11248    | CXPACKET              | 11110   | 0 . . .   Attached are the execution plans from SQL Server 2012 and SQL Server 2014.       Any help much appreciated. System installation is stalled on Compaq Presario R3000 How can I recreate geometry using a photograph? Converting a system command to python for file search and delete How to prove that there are finitely many $n$ such that $f(1+1/n) = \frac{1}{n+1}$? Let $f : \mathbb{C} \to \mathbb{C}$ be analytical. How does one prove that there are finitely many positive $n$ such that $f(1+1/n) = \frac{1}{n+1}$ ? A hint is given: think of the expression $\frac{z-1}{z}$.  I think this is a pretty hard homework assignment. Here's what I tried so far: Let $g(z) = \frac{z-1}{z} $ . Then $g(1 + 1/n) = \frac{1 + 1/n - 1}{ 1 + 1/n} = \frac{1/n}{1+1/n} = \frac{1}{n+1} $. So the given hint is probably indeed useful, but I'm not sure yet how.   Another thing I tried to use to obtain the solution was the given fact that $f$ is analytical. Let $f = g$. Then $f(z) = f(x+iy) = \frac{ x + iy - 1}{x+iy} = \frac{x+iy-1}{x+iy} \cdot \frac{x-iy}{x-iy} = \frac{x^2 - 2 i x y + y^2 + iy}{x^2 + y^2} = 1 + i \frac{y(1-2x)}{x^2 + y^2} $ . Now I have written $f$ in the form $ f = u + i v$, with $u,v$ being real-valued functions. So now we can use the Cauchy-Riemann Integral formulas: $ \frac{d}{dx} 1 = 0 = \frac{d}{dy} \frac{ y(1-2x) }{x^2+y^2} \implies \frac{y(1-2x)}{x^2+y^2} = c $. The other formula yields a similar result.   But now what? I still have no idea how to prove that there are finitely many positive $n$ such that the aforementioned equality holds. Should I use any of the results I already obtained? If so, how? If not, do you know how to prove this? Is "flu" the only disease usually accompanied with "the"? From what I understand, you don't usually use "the" with diseases / illnesses / disorders ("have Alzheimer's / diarrhea / claustrophobia / dyslexia, etc." not "have the..."), but it's perfectly normal to say "have the flu". Is the flu the only exception?   Edit (this is only a side part, and can be glossed over if you will, so it's not a duplicate imo): What's so significant about the flu that other conditions don't have, so that it's entitled to "the"?