How to sum binomial coefficients which are multiples of 3?
closed form for $\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+...+\binom{n}{n}$
Not sure if this option ever existed, but the new QGIS version hides a lot of things somewhere else and this seems so simple that I'm kind of convinced it was possible to do this before. I have quite a large data set with ship tracks. I now want to analyse individual ship tracks (made from point layer with  Plugin). However, it is so crowded that it is hard to look at individual tracks. I don't want to save each single track as a new layer, so I wonder if it is possible to select one track in the Attribute table and display only that one on the map/hide the others? How to prove $\sum_{k=1}^{n}F_k = F_{n+2}-1$ by induction when $F_n$ is the Fibonacci sequence Let $F_n$ be the Fibonacci sequence where  $F_0$ = 0  , $F_1$ = 1 and  $F_n$ = $F_{n-1}$ + $F_{n-2}$.                   I want to prove the following by induction. $$\sum_{k=1}^{n}F_k = F_{n+2}-1$$ Here is what I have so far.  Can anybody tell me if I am right?  Base case = n=1  $F_k$ = 1 = LHS  $F_{1+2}$ - 1 = 2 - 1 = 1 = RHS  Statement is true for n=1.  Assume statement is true for n=i.  $$\sum_{k=1}^{i}F_k = F_{i+2}-1$$  Prove statement is true for n=i+1.  $$\sum_{k=1}^{i+1}F_{k} = F_{((i+1)+2)}-1$$  $F_1$ + $F_2$ +...+ $F_i$ + $F_{i+1}$ = $F_{((i+1)+2)}-1$  $F_{i+2}$ - 1 + $F_{i+1}$ = $F_{((i+1)+2)}-1$  $F_{((i+1)+2)}-1$ = $F_{((i+1)+2)}-1$  Statement holds for n=i+1 Add different materials to different parts of a mesh? How do you add more than one material to a mesh?   I have tried adding two materials to the object, and then assigning one of the materials to a vertex-group, but I could not find a place to assign it.  I am making a striped bouncy ball that I want to deform, so adding a different material by separating the mesh into different objects would not be ideal. Is there a way to reach a `protected` member of another object from a derived type? class MyBase {     protected object PropertyOfBase { get; set; } }  class MyType : MyBase {     void MyMethod(MyBase parameter)     {         // I am looking for:         object p = parameter.PropertyOfBase;  // error CS1540: Cannot access protected member 'MyBase.PropertyOfBase' via a qualifier of type 'MyBase'; the qualifier must be of type 'MyType' (or derived from it)     } }   Is there a way to get a protected property of a parameter of a type from an extending type without reflection? Since the extending class knows of the property through its base type, it would make sense if possible. How to find all subgroups of a direct product? I am wondering how do we find all subgroups of a direct product? Is there a method to find it?  For example, how can we find all the subgroups of $\mathbb{Z}_2\times\mathbb{Z}_2$?  There is the answer: $\{(0, 0), (1, 0)\}, \{(0, 0), (0, 1)\}, \{(0, 0), (1, 1)\}$, and the improper and trivial subgroup. But I do not know the method how to find it.   All I can think about is the following: since the order of $\mathbb{Z}_2\times\mathbb{Z}_2$ is $4$, the order of a subgroup of it can be $1,2$ or $4$.   If order is $1$, then it is trivial, and if order is $4$, it is proper. Then, the order of a subgroup is $2$. We know all the elements of   $\mathbb{Z}_2\times\mathbb{Z}_2$, so we try all the pairs.   Is that the method?   But what if we are given a bigger order product, such as $\mathbb{Z}_2\times\mathbb{Z}_8\times\mathbb{Z}_3\times\mathbb{Z}_9$? It looks very long to try all possibilities.   Can anyone help?  Thanks Center of $D_6$ is $\mathbb{Z}_2$ The center of $D_6$ is isomorphic to $\mathbb{Z}_2$.   I have that $$D_6=\left&lt; a,b \mid a^6=b^2=e,\, ba=a^{-1}b\right&gt;$$ $$\Rightarrow D_6=\{e,a,a^2,a^3,a^4,a^5,b,ab,a^2b,a^3b,a^4b,a^5b\}.$$ My method for trying to do this has been just checking elements that could be candidates. I've widdled it down to that the only elements that commute with all of $D_6$ must be $\{e,a^3\}$ but I got there by finding a pair of elements that didn't commute for all other elements and I still haven't even shown that $a^3$ commutes with everything. For example, I have been trying to show now that $$a^3b=ba^3$$ and haven't gotten too far yet but if I had to answer a question like this on the exam, I feel it would be difficult, is there any kind of trick or hints other than brute force using the relations to get that $a^3$ commutes with everything?  For the solution once I have that the center of $D_6$ is what I think then as there is only one group of order $2$ up to isomorphism, it must be isomorphic to $\mathbb{Z}_2$.  Ideally a way that doesn't appeal to $D_6$ as symmetries of the hexagon if that seems possible. Take the following script:  #!/bin/sh sed 's/(127\.0\.1\.1)\s/\1/' [some file]   If I try to run this in sh (dash here), it'll fail because of the parentheses, which need to be escaped. But I don't need to escape the backslashes themselves (between the octets, or in the \s or \1). What's the rule here? What about when I need to use {...} or [...]? Is there a list of what I do and don't need to escape? What is the proper behavior when someone abuses the close-vote superpowers? When documenting a style file, it would be nice to be able to easily put side by side different examples of code, side-by-side with the output they produce.  What recommendations to people have to do this efficiently, and &aelig;sthetically? Why aren't logarithms defined for negative $x$? Given a logarithm is true, if and only if, $y = \log_b{x}$ and $b^y = x$ (and $x$ and $b$ are positive, and $b$ is not equal to $1$), are true, why aren't logarithms defined for negative numbers?  Why can't $b$ be negative? Take $(-2)^3 = -8$ for example. Turning that into a logarithm, we get $3 = \log_{(-2)}{(-8)}$ which is an invalid equation as $b$ and $x$ are not positive! Why is this? Why do objects appear smaller when farther away? Why is my USB drive showing corrupted data when plugged as an internal SATA drive? If Newton's third law is true, why can we sink in sand? Newton's third law of motion states that every action has an equal and opposite reaction. That is the reason we do not sink into the earth, because when our weight exerts a force on the earth it also exerts an equal and opposite force on us.  But when we stand on quicksand or on fluids we can sink in. How is this possible? Does it not exert an equal and opposite force on us? Or are Newton's laws different in the case of fluids and substances of low densities? Looking for Space Exploration Book which is collection of different sub-stories The Book is about an earthship exploring the frontiers that I read as a kid (80's). All the stories are told by the same character but different encounters and events.    The first story is about the crew exploring the planet in a number of separate ships when one of them is taken over by a telepathic alien. As the ship returns however the surprise is sprung on the alien that the Mother ship is Computer controlled. Another is the ship shrinking and then finding after its shrunk beyond the atom that it comes back in as gigantic compared to the Universe in a closed loop.   Hopefully that's enough to trigger a memory. What is the exact use of the symbols $\partial$, $\delta$ and $\mathrm{d}$ in derivatives in physics? How are they different and when are they used? It would be nice to get that settled once and for all.  $$\frac{\partial y}{\partial x}, \frac{\delta y}{\delta x}, \frac{\mathrm{d} y}{\mathrm{d} x}$$   For what I know, $\mathrm{d}$ is used as a small infinitisemal change (and I guess the straight-up letter $\mathrm{d}$ is usual notation instead of italic $d$, simply to tell the difference from a variable). Of course we also have the big delta $\Delta$ to describe a finite (non-negligible) difference. And I have some vague idea that $\partial$ is used for partial derivatives in case of e.g. three-dimensional variables. Same goes for $\delta$, which I would have sworn was the same as $\partial$ until reading this answer on Math.SE:    Then to make the confusion total I noticed an equation like $\delta Q=\mathrm{d}U+\delta W$ and read in a physics text book that:     The fact that the amount of heat [added between two states] is dependent on the path is indicated by the symbol $\delta$...   So it seems $\delta$ means something more? The text book continues and says that:     a function [like the change in internal energy] is called a state function and its change is indicated by the symbol $\mathrm{d}$...   Here I am unsure of exactly why a $\mathrm d$ refers to a state function.   So to sum it up: down to the bone of it, what is $\delta$, $\partial$ and $\mathrm{d}$ exactly, when we are talking derivatives in physics.  Addition  Especially when reading a mathematical process on a physical equation like this procedure:  $$\delta Q=\mathrm{d}U+p\mathrm{d}V \Rightarrow\\ Q=\Delta U+\int_1^2 p \mathrm{d}V$$  It appears that $\delta$ and $\mathrm{d}$ are the same thing. An integral operation handles it the same way apparently? What are the pros and cons of adding a vapor barrier behind shower backer board? After much research, I have a lot of conflicting advise about the need to add a vapor barrier before installing the backer (hardy / cement) board for my bathroom shower. The &quot;expert&quot; at Home Depot, and  seem to think this is a very good idea, and even suggest sealing it with roofing tape.  (The guy on Youtube tapes it directly to the sill of tub.) However, when I asked about the tape and how far down I should go at Lowes, they told me they'd never hear of anyone doing that, and that it didn't seem like a good idea.   Likewise,  putting down caulking on the sill before applying the backer board, and then again after before adding tile, and then caulking a 3rd time under the tiles before grouting. Should I add a vapor barrier between the frame and the cement board?  Why or why not? Why are there so many different ways to measure disk usage? When I sum up the sizes of my files, I get one figure. If I run du, I get another figure. If I run du on all the files on my partition, it doesn't match what df claims is used. Why are there so many different figures for the total size of my files? Can't computers add?  Speaking of adding: when I add the “Used” and “Available” columns of df, I don't get the total figure. And that total figure is smaller than the size of my partition. And if I add up my partition sizes I don't get my disk size! What gives? Server 2008 Net Send or Msg? I have a Windows 2008 server running as a domain controller. The details are as follows:  Domain: corp Server Name: oxygen   The workstations are all running Windows XP SP3 and have fixed IP addresses.  I am trying to use the net send functionality to be able to send simple messages to either individual machines/users and also to all machines. This is so I can send a message to a user to alert them about something, or likewise alert everyone of something such as the server being rebooted.  Is this possible using Windows Server 2008? If so, what would the command be? The following question is taken from Royden's Real Analysis $4$th edition, Chapter $3,$ question $28,$ page $67:$     Question: Show that Egoroff's Theorem continues to hold if the convergence is pointwise a.e. and $f$ is finite a.e., that is.       Assume $E$ has finite measure. Let $\{f_n\}$ be a sequence of measurable functions on $E$ that converges pointwise almost everywhere on $E$ to the real-valued function $f$ which is finite almost everywhere. Then for each $\varepsilon&gt;0,$ there is a closed set $F$ contained in $E$ for which    $$\{f_n\}\to f \text{ uniformly on }F \text{ and }m(E\setminus F)&lt;\varepsilon.$$   My attempt:   Let $A = \{ x\in E: f_n\not\to f \text{ pointwise} \}$ and $B=\{ x\in E: |f|=\infty \}.$  By assumption,  $$m(A)=m(B)=0.$$ So $A$ and $B$ are measurable.  Note that  $$m[E\setminus (A\cup B)] = m(E) - m(A\cup B) = m(E).$$ Fix $\varepsilon&gt;0.$ Since $\{f_n\}$ converges to $f$ pointwise on $E\setminus (A\cup B)$ to the real-valued function $f,$ by Egoroff's Theorem, there exists a closed set $F$ contained in $(E\setminus (A\cup B))\subseteq E$ for which  $$f_n\to f \text{ uniformly on } F \text{ and }m[(E\setminus(A\cup B)) \setminus F] &lt; \varepsilon.$$ Since $E$ has finite measure and $F\subseteq E,$ by monotonicity,  $F$ has finite measure. As $F$ is closed, it is also measurable.  Therefore, by Excision property, we have $$m(E\setminus F) = m(E) - m(F) = m[E\setminus(A\cup B)] - m(F) = m[(E\setminus(A\cup B)) \setminus F]&lt;\varepsilon.$$  Is my proof correct? I ask for verification because in this  Kenny's comment about countable union of null sets. I do not use this anywhere in my proof. So I wonder whether my proof miss out something. MWE:  \documentclass{article}  \usepackage{listings}  \begin{document}  IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII hardware/software  \end{document}   This results in this output:    If \usepackage{listings} is commented out, the hyphenation rules return to normal:    And just to note, in the actual environment I'm working in, it will also hyphenate between hard•ware, while with listings enabled, it hyphenates hardware/•software, as in the first image:    Two questions:   Why is this happening? Is there a way to disable this? I don't really want this behaviour just because I'm typing code occasionally in a document.