Prove that every geometric sequence allows for $S_n(S_{3n}-S_{2n})=(S_{2n}-S_{n})^2$












3












$begingroup$



$S_n(S_{3n}-S_{2n})=(S_{2n}-S_{n})^2$




Is there a way to prove this without expanding everything based on the geometric sum formula? I get lost very easily when trying to solve this conventionally and I feel that I am missing an obvious solution.






sequences-and-series geometric-series







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$endgroup$

















    3












    $begingroup$



    $S_n(S_{3n}-S_{2n})=(S_{2n}-S_{n})^2$




    Is there a way to prove this without expanding everything based on the geometric sum formula? I get lost very easily when trying to solve this conventionally and I feel that I am missing an obvious solution.






    sequences-and-series geometric-series







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$



      $S_n(S_{3n}-S_{2n})=(S_{2n}-S_{n})^2$




      Is there a way to prove this without expanding everything based on the geometric sum formula? I get lost very easily when trying to solve this conventionally and I feel that I am missing an obvious solution.






      sequences-and-series geometric-series







      share|cite|improve this question









      $endgroup$





      $S_n(S_{3n}-S_{2n})=(S_{2n}-S_{n})^2$




      Is there a way to prove this without expanding everything based on the geometric sum formula? I get lost very easily when trying to solve this conventionally and I feel that I am missing an obvious solution.






      sequences-and-series geometric-series




      sequences-and-series geometric-series






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 3 at 19:23









      daedsidogdaedsidog

      30417




      30417






















          2 Answers
          2






          active

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          5












          $begingroup$

          Write



          $$S_n = a sum_{k=0}^{n-1} r^k$$



          Then



          $$S_{2 n} = S_n + r^n S_n$$
          $$S_{3 n} = S_{2 n} + r^{2 n}S_n$$



          The above result follows from simple algebra.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is perfect. Thank you.
            $endgroup$
            – daedsidog
            Jan 3 at 19:46



















          3












          $begingroup$

          firsly:
          $$S_n=a_1frac{1-r^n}{1-r}$$
          so we can say that:
          $$S_nleft(S_{3n}-S_{2n}right)=a_1^2frac{1-r^n}{1-r}left(frac{(1-r^{3n})-(1-r^{2n})}{1-r}right)=a_1^2frac{(1-r^n)(r^{2n}-r^{3n})}{(1-r)^2}$$
          and:
          $$(S_{2n}-S_n)^2=left[a_1frac{1-r^{2n}}{(1-r)}-a_1frac{1-r^n}{(1-r)}right]^2=left[a_1frac{r^n-r^{2n}}{(1-r)}right]^2=a_1^2frac{(r^n-r^{2n})^2}{(1-r)^2}$$
          now we just need to show that the top of the fractions are equivalent:
          $$(r^n-r^{2n})^2=(r^n)^2-2r^nr^{2n}+(r^{2n})^2=r^{2n}-2r^{3n}+r^{4n}$$
          $$(1-r^n)(r^{2n}-r^{3n})=r^{2n}-r^{3n}-r^{3n}+r^{4n}=r^{2n}-2r^{3n}+r^{4n}$$
          so the two are equivalent






          share|cite|improve this answer









          $endgroup$














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            2 Answers
            2






            active

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            2 Answers
            2






            active

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            active

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            active

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            5












            $begingroup$

            Write



            $$S_n = a sum_{k=0}^{n-1} r^k$$



            Then



            $$S_{2 n} = S_n + r^n S_n$$
            $$S_{3 n} = S_{2 n} + r^{2 n}S_n$$



            The above result follows from simple algebra.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              This is perfect. Thank you.
              $endgroup$
              – daedsidog
              Jan 3 at 19:46
















            5












            $begingroup$

            Write



            $$S_n = a sum_{k=0}^{n-1} r^k$$



            Then



            $$S_{2 n} = S_n + r^n S_n$$
            $$S_{3 n} = S_{2 n} + r^{2 n}S_n$$



            The above result follows from simple algebra.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              This is perfect. Thank you.
              $endgroup$
              – daedsidog
              Jan 3 at 19:46














            5












            5








            5





            $begingroup$

            Write



            $$S_n = a sum_{k=0}^{n-1} r^k$$



            Then



            $$S_{2 n} = S_n + r^n S_n$$
            $$S_{3 n} = S_{2 n} + r^{2 n}S_n$$



            The above result follows from simple algebra.






            share|cite|improve this answer









            $endgroup$



            Write



            $$S_n = a sum_{k=0}^{n-1} r^k$$



            Then



            $$S_{2 n} = S_n + r^n S_n$$
            $$S_{3 n} = S_{2 n} + r^{2 n}S_n$$



            The above result follows from simple algebra.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 3 at 19:36









            Ron GordonRon Gordon

            123k14156267




            123k14156267












            • $begingroup$
              This is perfect. Thank you.
              $endgroup$
              – daedsidog
              Jan 3 at 19:46


















            • $begingroup$
              This is perfect. Thank you.
              $endgroup$
              – daedsidog
              Jan 3 at 19:46
















            $begingroup$
            This is perfect. Thank you.
            $endgroup$
            – daedsidog
            Jan 3 at 19:46




            $begingroup$
            This is perfect. Thank you.
            $endgroup$
            – daedsidog
            Jan 3 at 19:46











            3












            $begingroup$

            firsly:
            $$S_n=a_1frac{1-r^n}{1-r}$$
            so we can say that:
            $$S_nleft(S_{3n}-S_{2n}right)=a_1^2frac{1-r^n}{1-r}left(frac{(1-r^{3n})-(1-r^{2n})}{1-r}right)=a_1^2frac{(1-r^n)(r^{2n}-r^{3n})}{(1-r)^2}$$
            and:
            $$(S_{2n}-S_n)^2=left[a_1frac{1-r^{2n}}{(1-r)}-a_1frac{1-r^n}{(1-r)}right]^2=left[a_1frac{r^n-r^{2n}}{(1-r)}right]^2=a_1^2frac{(r^n-r^{2n})^2}{(1-r)^2}$$
            now we just need to show that the top of the fractions are equivalent:
            $$(r^n-r^{2n})^2=(r^n)^2-2r^nr^{2n}+(r^{2n})^2=r^{2n}-2r^{3n}+r^{4n}$$
            $$(1-r^n)(r^{2n}-r^{3n})=r^{2n}-r^{3n}-r^{3n}+r^{4n}=r^{2n}-2r^{3n}+r^{4n}$$
            so the two are equivalent






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              firsly:
              $$S_n=a_1frac{1-r^n}{1-r}$$
              so we can say that:
              $$S_nleft(S_{3n}-S_{2n}right)=a_1^2frac{1-r^n}{1-r}left(frac{(1-r^{3n})-(1-r^{2n})}{1-r}right)=a_1^2frac{(1-r^n)(r^{2n}-r^{3n})}{(1-r)^2}$$
              and:
              $$(S_{2n}-S_n)^2=left[a_1frac{1-r^{2n}}{(1-r)}-a_1frac{1-r^n}{(1-r)}right]^2=left[a_1frac{r^n-r^{2n}}{(1-r)}right]^2=a_1^2frac{(r^n-r^{2n})^2}{(1-r)^2}$$
              now we just need to show that the top of the fractions are equivalent:
              $$(r^n-r^{2n})^2=(r^n)^2-2r^nr^{2n}+(r^{2n})^2=r^{2n}-2r^{3n}+r^{4n}$$
              $$(1-r^n)(r^{2n}-r^{3n})=r^{2n}-r^{3n}-r^{3n}+r^{4n}=r^{2n}-2r^{3n}+r^{4n}$$
              so the two are equivalent






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                firsly:
                $$S_n=a_1frac{1-r^n}{1-r}$$
                so we can say that:
                $$S_nleft(S_{3n}-S_{2n}right)=a_1^2frac{1-r^n}{1-r}left(frac{(1-r^{3n})-(1-r^{2n})}{1-r}right)=a_1^2frac{(1-r^n)(r^{2n}-r^{3n})}{(1-r)^2}$$
                and:
                $$(S_{2n}-S_n)^2=left[a_1frac{1-r^{2n}}{(1-r)}-a_1frac{1-r^n}{(1-r)}right]^2=left[a_1frac{r^n-r^{2n}}{(1-r)}right]^2=a_1^2frac{(r^n-r^{2n})^2}{(1-r)^2}$$
                now we just need to show that the top of the fractions are equivalent:
                $$(r^n-r^{2n})^2=(r^n)^2-2r^nr^{2n}+(r^{2n})^2=r^{2n}-2r^{3n}+r^{4n}$$
                $$(1-r^n)(r^{2n}-r^{3n})=r^{2n}-r^{3n}-r^{3n}+r^{4n}=r^{2n}-2r^{3n}+r^{4n}$$
                so the two are equivalent






                share|cite|improve this answer









                $endgroup$



                firsly:
                $$S_n=a_1frac{1-r^n}{1-r}$$
                so we can say that:
                $$S_nleft(S_{3n}-S_{2n}right)=a_1^2frac{1-r^n}{1-r}left(frac{(1-r^{3n})-(1-r^{2n})}{1-r}right)=a_1^2frac{(1-r^n)(r^{2n}-r^{3n})}{(1-r)^2}$$
                and:
                $$(S_{2n}-S_n)^2=left[a_1frac{1-r^{2n}}{(1-r)}-a_1frac{1-r^n}{(1-r)}right]^2=left[a_1frac{r^n-r^{2n}}{(1-r)}right]^2=a_1^2frac{(r^n-r^{2n})^2}{(1-r)^2}$$
                now we just need to show that the top of the fractions are equivalent:
                $$(r^n-r^{2n})^2=(r^n)^2-2r^nr^{2n}+(r^{2n})^2=r^{2n}-2r^{3n}+r^{4n}$$
                $$(1-r^n)(r^{2n}-r^{3n})=r^{2n}-r^{3n}-r^{3n}+r^{4n}=r^{2n}-2r^{3n}+r^{4n}$$
                so the two are equivalent







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 3 at 19:36









                Henry LeeHenry Lee

                2,189319




                2,189319






























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