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












3















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




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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    3















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




    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

























      3












      3








      3








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




      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














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




      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











      share|cite|improve this question




      share|cite|improve this question










      asked 2 days ago









      daedsidog

      22916




      22916






















          2 Answers
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          oldest

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          5














          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





















          • This is perfect. Thank you.
            – daedsidog
            2 days ago



















          3














          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





















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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5














            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





















            • This is perfect. Thank you.
              – daedsidog
              2 days ago
















            5














            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





















            • This is perfect. Thank you.
              – daedsidog
              2 days ago














            5












            5








            5






            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












            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 2 days ago









            Ron Gordon

            122k14154263




            122k14154263












            • This is perfect. Thank you.
              – daedsidog
              2 days ago


















            • This is perfect. Thank you.
              – daedsidog
              2 days ago
















            This is perfect. Thank you.
            – daedsidog
            2 days ago




            This is perfect. Thank you.
            – daedsidog
            2 days ago











            3














            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


























              3














              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
























                3












                3








                3






                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












                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 2 days ago









                Henry Lee

                1,773218




                1,773218






























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