Sum of squares in geometric progression












5












$begingroup$



In the geometric progression $b_1, b_2, b_3,ldots, b_1+b_3+b_5=10$
and $b_2+b_4=5$. Find the sum of the squares of the first five terms.




If you solve for the first term and the common ratio, you will get nasty radicals and then squaring and adding is not a good option. How do I do it then?






algebra-precalculus geometric-progressions







share|cite|improve this question











$endgroup$

















    5












    $begingroup$



    In the geometric progression $b_1, b_2, b_3,ldots, b_1+b_3+b_5=10$
    and $b_2+b_4=5$. Find the sum of the squares of the first five terms.




    If you solve for the first term and the common ratio, you will get nasty radicals and then squaring and adding is not a good option. How do I do it then?






    algebra-precalculus geometric-progressions







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      1



      $begingroup$



      In the geometric progression $b_1, b_2, b_3,ldots, b_1+b_3+b_5=10$
      and $b_2+b_4=5$. Find the sum of the squares of the first five terms.




      If you solve for the first term and the common ratio, you will get nasty radicals and then squaring and adding is not a good option. How do I do it then?






      algebra-precalculus geometric-progressions







      share|cite|improve this question











      $endgroup$





      In the geometric progression $b_1, b_2, b_3,ldots, b_1+b_3+b_5=10$
      and $b_2+b_4=5$. Find the sum of the squares of the first five terms.




      If you solve for the first term and the common ratio, you will get nasty radicals and then squaring and adding is not a good option. How do I do it then?






      algebra-precalculus geometric-progressions




      algebra-precalculus geometric-progressions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 2 '15 at 6:18









      nbubis

      27.4k553110




      27.4k553110










      asked Mar 2 '15 at 4:44







      user167045





























          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.



          More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
            $endgroup$
            – user167045
            Mar 2 '15 at 5:04












          • $begingroup$
            @Garvil I don't see any leftover terms.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:10










          • $begingroup$
            Yeah, sorry. My bad.
            $endgroup$
            – user167045
            Mar 2 '15 at 5:12










          • $begingroup$
            No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:13










          • $begingroup$
            Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
            $endgroup$
            – Janko Bracic
            Mar 2 '15 at 5:27



















          2












          $begingroup$

          Let $b_1=a$, and the common ratio be $r$. Given that
          $$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
          What is the value of
          $$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
          Realize that:
          $$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
          simplifies to the required sum. On substituting:
          $$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:09












          • $begingroup$
            20 terms cancel out; remaining five are the required five.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:11












          Your Answer








          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%2f1171397%2fsum-of-squares-in-geometric-progression%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown
























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.



          More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
            $endgroup$
            – user167045
            Mar 2 '15 at 5:04












          • $begingroup$
            @Garvil I don't see any leftover terms.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:10










          • $begingroup$
            Yeah, sorry. My bad.
            $endgroup$
            – user167045
            Mar 2 '15 at 5:12










          • $begingroup$
            No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:13










          • $begingroup$
            Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
            $endgroup$
            – Janko Bracic
            Mar 2 '15 at 5:27
















          6












          $begingroup$

          Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.



          More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
            $endgroup$
            – user167045
            Mar 2 '15 at 5:04












          • $begingroup$
            @Garvil I don't see any leftover terms.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:10










          • $begingroup$
            Yeah, sorry. My bad.
            $endgroup$
            – user167045
            Mar 2 '15 at 5:12










          • $begingroup$
            No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:13










          • $begingroup$
            Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
            $endgroup$
            – Janko Bracic
            Mar 2 '15 at 5:27














          6












          6








          6





          $begingroup$

          Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.



          More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.






          share|cite|improve this answer











          $endgroup$



          Hint: $(b_1+b_3+b_5)^2-(b_2+b_4)^2$.



          More suggestions: $(x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz$, $(x+y)^2=x^2+2xy+y^2$ (this you know, I guess), now use the fact that in geometric progression one has $b_{n-1}b_{n+1}=b_{n}^{2}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 2 '15 at 5:08

























          answered Mar 2 '15 at 4:56









          Janko BracicJanko Bracic

          2,783613




          2,783613












          • $begingroup$
            After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
            $endgroup$
            – user167045
            Mar 2 '15 at 5:04












          • $begingroup$
            @Garvil I don't see any leftover terms.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:10










          • $begingroup$
            Yeah, sorry. My bad.
            $endgroup$
            – user167045
            Mar 2 '15 at 5:12










          • $begingroup$
            No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:13










          • $begingroup$
            Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
            $endgroup$
            – Janko Bracic
            Mar 2 '15 at 5:27


















          • $begingroup$
            After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
            $endgroup$
            – user167045
            Mar 2 '15 at 5:04












          • $begingroup$
            @Garvil I don't see any leftover terms.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:10










          • $begingroup$
            Yeah, sorry. My bad.
            $endgroup$
            – user167045
            Mar 2 '15 at 5:12










          • $begingroup$
            No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:13










          • $begingroup$
            Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
            $endgroup$
            – Janko Bracic
            Mar 2 '15 at 5:27
















          $begingroup$
          After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
          $endgroup$
          – user167045
          Mar 2 '15 at 5:04






          $begingroup$
          After solving what you gave, I am left with $50=b_1^2+b_2^2+b_3^2+b_4^2+b_5^2-2b_5$
          $endgroup$
          – user167045
          Mar 2 '15 at 5:04














          $begingroup$
          @Garvil I don't see any leftover terms.
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:10




          $begingroup$
          @Garvil I don't see any leftover terms.
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:10












          $begingroup$
          Yeah, sorry. My bad.
          $endgroup$
          – user167045
          Mar 2 '15 at 5:12




          $begingroup$
          Yeah, sorry. My bad.
          $endgroup$
          – user167045
          Mar 2 '15 at 5:12












          $begingroup$
          No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:13




          $begingroup$
          No problem. And damn... I never get accepted answers because I'm always the second to post the answer :D Thanks for the upvote. Will edit when possible.
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:13












          $begingroup$
          Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
          $endgroup$
          – Janko Bracic
          Mar 2 '15 at 5:27




          $begingroup$
          Garvil, can you change the accepted answer, please? Let us support enthusiastic young people :)
          $endgroup$
          – Janko Bracic
          Mar 2 '15 at 5:27











          2












          $begingroup$

          Let $b_1=a$, and the common ratio be $r$. Given that
          $$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
          What is the value of
          $$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
          Realize that:
          $$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
          simplifies to the required sum. On substituting:
          $$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:09












          • $begingroup$
            20 terms cancel out; remaining five are the required five.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:11
















          2












          $begingroup$

          Let $b_1=a$, and the common ratio be $r$. Given that
          $$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
          What is the value of
          $$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
          Realize that:
          $$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
          simplifies to the required sum. On substituting:
          $$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:09












          • $begingroup$
            20 terms cancel out; remaining five are the required five.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:11














          2












          2








          2





          $begingroup$

          Let $b_1=a$, and the common ratio be $r$. Given that
          $$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
          What is the value of
          $$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
          Realize that:
          $$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
          simplifies to the required sum. On substituting:
          $$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$






          share|cite|improve this answer











          $endgroup$



          Let $b_1=a$, and the common ratio be $r$. Given that
          $$a+ar^2+ar^4=10 qquadtext{and}qquad ar+ar^3=5.$$
          What is the value of
          $$(a)^2 +(ar)^2 +(ar^2)^2 +(ar^3)^2 +(ar^4)^2?$$
          Realize that:
          $$(a+ar^2+ar^4)^2 -(ar+ar^3)^2$$
          simplifies to the required sum. On substituting:
          $$(10)^2 -(5)^2 = (10+5)(10-5) = 75.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 30 '18 at 1:41









          Rócherz

          3,0263823




          3,0263823










          answered Mar 2 '15 at 5:08









          KugelblitzKugelblitz

          3,3491449




          3,3491449












          • $begingroup$
            Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:09












          • $begingroup$
            20 terms cancel out; remaining five are the required five.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:11


















          • $begingroup$
            Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:09












          • $begingroup$
            20 terms cancel out; remaining five are the required five.
            $endgroup$
            – Kugelblitz
            Mar 2 '15 at 5:11
















          $begingroup$
          Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:09






          $begingroup$
          Note: Janko seems to have already given you the big hint. And no, it simplifies properly into those five terms. @Garvil
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:09














          $begingroup$
          20 terms cancel out; remaining five are the required five.
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:11




          $begingroup$
          20 terms cancel out; remaining five are the required five.
          $endgroup$
          – Kugelblitz
          Mar 2 '15 at 5:11


















          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%2f1171397%2fsum-of-squares-in-geometric-progression%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

          How to change which sound is reproduced for terminal bell?

          Image
          3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
          Read more

          Can I use Tabulator js library in my java Spring + Thymeleaf project?

          Image
          0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
          Read more

          Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

          Image
          up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
          Read more