Sum of a combination












0












$begingroup$


i cant find an expression for:



$sum_{k=0}^{lfloor
frac{p}{2}rfloor}$ ${p}choose{k}$



The hint given by the exercise is to do it by seperating the cases where p is odd and even.



What I did so far:
When $p$ is even, $lfloor p/2 rfloor = p/2$, so all we need is find an expression
$sum_{k=0}^{ frac{p}{2}}$ ${p}choose{k}$, which I was not able to do.



When $p$ is odd, there is $q$ so that $p=2q+1$ and $p/2= q +1/2$ which means $lfloor p/2 rfloor = q$, and I really dont know what to do with this.



The previous question was to show that functions that are of the form $ax+b$ are both convex and concave.



I thought that I should try to put some values in to look for a pattern and here is what I have :
for $p$ even :



$p=2, sum_{k=0}^{1}$ ${2}choose{k}$$=1+2=3$



$p=4, sum_{k=0}^{2}$ ${4}choose{k}$$=1+6+4=11$



$p=6, sum_{k=0}^{3}$ ${6}choose{k}$$=1+6+15+20=42$



no apparent pattern.



when p is odd,



$p=1, sum_{k=0}^{0}$ ${1}choose{k}$$=1$



$p=3, sum_{k=0}^{1}$ ${3}choose{k}$$=1+3=4$



$p=5, sum_{k=0}^{2}$ ${5}choose{k}$$=1+5+10=16$



$p=7, sum_{k=0}^{4}$ ${7}choose{k}$$=1+7+21+35+35=99$



No apparent pattern here



Thank you for your help






summation combinations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Your calculation for $p=7$ is wrong.
    $endgroup$
    – Robert Israel
    Jan 13 at 21:14
















0












$begingroup$


i cant find an expression for:



$sum_{k=0}^{lfloor
frac{p}{2}rfloor}$ ${p}choose{k}$



The hint given by the exercise is to do it by seperating the cases where p is odd and even.



What I did so far:
When $p$ is even, $lfloor p/2 rfloor = p/2$, so all we need is find an expression
$sum_{k=0}^{ frac{p}{2}}$ ${p}choose{k}$, which I was not able to do.



When $p$ is odd, there is $q$ so that $p=2q+1$ and $p/2= q +1/2$ which means $lfloor p/2 rfloor = q$, and I really dont know what to do with this.



The previous question was to show that functions that are of the form $ax+b$ are both convex and concave.



I thought that I should try to put some values in to look for a pattern and here is what I have :
for $p$ even :



$p=2, sum_{k=0}^{1}$ ${2}choose{k}$$=1+2=3$



$p=4, sum_{k=0}^{2}$ ${4}choose{k}$$=1+6+4=11$



$p=6, sum_{k=0}^{3}$ ${6}choose{k}$$=1+6+15+20=42$



no apparent pattern.



when p is odd,



$p=1, sum_{k=0}^{0}$ ${1}choose{k}$$=1$



$p=3, sum_{k=0}^{1}$ ${3}choose{k}$$=1+3=4$



$p=5, sum_{k=0}^{2}$ ${5}choose{k}$$=1+5+10=16$



$p=7, sum_{k=0}^{4}$ ${7}choose{k}$$=1+7+21+35+35=99$



No apparent pattern here



Thank you for your help






summation combinations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Your calculation for $p=7$ is wrong.
    $endgroup$
    – Robert Israel
    Jan 13 at 21:14














0












0








0





$begingroup$


i cant find an expression for:



$sum_{k=0}^{lfloor
frac{p}{2}rfloor}$ ${p}choose{k}$



The hint given by the exercise is to do it by seperating the cases where p is odd and even.



What I did so far:
When $p$ is even, $lfloor p/2 rfloor = p/2$, so all we need is find an expression
$sum_{k=0}^{ frac{p}{2}}$ ${p}choose{k}$, which I was not able to do.



When $p$ is odd, there is $q$ so that $p=2q+1$ and $p/2= q +1/2$ which means $lfloor p/2 rfloor = q$, and I really dont know what to do with this.



The previous question was to show that functions that are of the form $ax+b$ are both convex and concave.



I thought that I should try to put some values in to look for a pattern and here is what I have :
for $p$ even :



$p=2, sum_{k=0}^{1}$ ${2}choose{k}$$=1+2=3$



$p=4, sum_{k=0}^{2}$ ${4}choose{k}$$=1+6+4=11$



$p=6, sum_{k=0}^{3}$ ${6}choose{k}$$=1+6+15+20=42$



no apparent pattern.



when p is odd,



$p=1, sum_{k=0}^{0}$ ${1}choose{k}$$=1$



$p=3, sum_{k=0}^{1}$ ${3}choose{k}$$=1+3=4$



$p=5, sum_{k=0}^{2}$ ${5}choose{k}$$=1+5+10=16$



$p=7, sum_{k=0}^{4}$ ${7}choose{k}$$=1+7+21+35+35=99$



No apparent pattern here



Thank you for your help






summation combinations







share|cite|improve this question









$endgroup$




i cant find an expression for:



$sum_{k=0}^{lfloor
frac{p}{2}rfloor}$ ${p}choose{k}$



The hint given by the exercise is to do it by seperating the cases where p is odd and even.



What I did so far:
When $p$ is even, $lfloor p/2 rfloor = p/2$, so all we need is find an expression
$sum_{k=0}^{ frac{p}{2}}$ ${p}choose{k}$, which I was not able to do.



When $p$ is odd, there is $q$ so that $p=2q+1$ and $p/2= q +1/2$ which means $lfloor p/2 rfloor = q$, and I really dont know what to do with this.



The previous question was to show that functions that are of the form $ax+b$ are both convex and concave.



I thought that I should try to put some values in to look for a pattern and here is what I have :
for $p$ even :



$p=2, sum_{k=0}^{1}$ ${2}choose{k}$$=1+2=3$



$p=4, sum_{k=0}^{2}$ ${4}choose{k}$$=1+6+4=11$



$p=6, sum_{k=0}^{3}$ ${6}choose{k}$$=1+6+15+20=42$



no apparent pattern.



when p is odd,



$p=1, sum_{k=0}^{0}$ ${1}choose{k}$$=1$



$p=3, sum_{k=0}^{1}$ ${3}choose{k}$$=1+3=4$



$p=5, sum_{k=0}^{2}$ ${5}choose{k}$$=1+5+10=16$



$p=7, sum_{k=0}^{4}$ ${7}choose{k}$$=1+7+21+35+35=99$



No apparent pattern here



Thank you for your help






summation combinations




summation combinations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 13 at 20:52









user310148user310148

43




43








  • 1




    $begingroup$
    Your calculation for $p=7$ is wrong.
    $endgroup$
    – Robert Israel
    Jan 13 at 21:14














  • 1




    $begingroup$
    Your calculation for $p=7$ is wrong.
    $endgroup$
    – Robert Israel
    Jan 13 at 21:14








1




1




$begingroup$
Your calculation for $p=7$ is wrong.
$endgroup$
– Robert Israel
Jan 13 at 21:14




$begingroup$
Your calculation for $p=7$ is wrong.
$endgroup$
– Robert Israel
Jan 13 at 21:14










1 Answer
1






active

oldest

votes


















1












$begingroup$

We have ${p choose k} = {p choose p-k}$, so
$$ sum_{k=0}^{lfloor p/2 rfloor} {p choose k} = sum_{k = p - lfloor p/2 rfloor}^p {p choose k}$$
Thus if $p$ is odd, ${0, ldots, lfloor p/2 rfloor }$ and ${p - lfloor p/2 rfloor, ldots, p}$ are disjoint and their union is ${0, ldots, p}$, so using the binomial theorem your sum is $2^{p-1}$.
If $p$ is even, $p/2$ is in both sets, so your sum is $2^{p-1} + {p choose p/2}/2$.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3072500%2fsum-of-a-combination%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    We have ${p choose k} = {p choose p-k}$, so
    $$ sum_{k=0}^{lfloor p/2 rfloor} {p choose k} = sum_{k = p - lfloor p/2 rfloor}^p {p choose k}$$
    Thus if $p$ is odd, ${0, ldots, lfloor p/2 rfloor }$ and ${p - lfloor p/2 rfloor, ldots, p}$ are disjoint and their union is ${0, ldots, p}$, so using the binomial theorem your sum is $2^{p-1}$.
    If $p$ is even, $p/2$ is in both sets, so your sum is $2^{p-1} + {p choose p/2}/2$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      We have ${p choose k} = {p choose p-k}$, so
      $$ sum_{k=0}^{lfloor p/2 rfloor} {p choose k} = sum_{k = p - lfloor p/2 rfloor}^p {p choose k}$$
      Thus if $p$ is odd, ${0, ldots, lfloor p/2 rfloor }$ and ${p - lfloor p/2 rfloor, ldots, p}$ are disjoint and their union is ${0, ldots, p}$, so using the binomial theorem your sum is $2^{p-1}$.
      If $p$ is even, $p/2$ is in both sets, so your sum is $2^{p-1} + {p choose p/2}/2$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        We have ${p choose k} = {p choose p-k}$, so
        $$ sum_{k=0}^{lfloor p/2 rfloor} {p choose k} = sum_{k = p - lfloor p/2 rfloor}^p {p choose k}$$
        Thus if $p$ is odd, ${0, ldots, lfloor p/2 rfloor }$ and ${p - lfloor p/2 rfloor, ldots, p}$ are disjoint and their union is ${0, ldots, p}$, so using the binomial theorem your sum is $2^{p-1}$.
        If $p$ is even, $p/2$ is in both sets, so your sum is $2^{p-1} + {p choose p/2}/2$.






        share|cite|improve this answer









        $endgroup$



        We have ${p choose k} = {p choose p-k}$, so
        $$ sum_{k=0}^{lfloor p/2 rfloor} {p choose k} = sum_{k = p - lfloor p/2 rfloor}^p {p choose k}$$
        Thus if $p$ is odd, ${0, ldots, lfloor p/2 rfloor }$ and ${p - lfloor p/2 rfloor, ldots, p}$ are disjoint and their union is ${0, ldots, p}$, so using the binomial theorem your sum is $2^{p-1}$.
        If $p$ is even, $p/2$ is in both sets, so your sum is $2^{p-1} + {p choose p/2}/2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 13 at 21:13









        Robert IsraelRobert Israel

        328k23216469




        328k23216469






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3072500%2fsum-of-a-combination%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Human spaceflight

            Image
            Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
            Read more

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            Image
            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
            Read more

            File:DeusFollowingSea.jpg

            Image
            File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
            Read more