Find general formula for the termsDeriving a formula to find the sum of a series.Find general formula for a seriesHow to find the general formula for this recursive problem?Demonstration of a simple formulaFinding general formula for sequenceGeneral formula for harmonic sequenceHow can I tell if the sequence re-cycles?The general term of such a recursion formulaFind an explicit formula for the recursive formulaThe general nth derivative formula of $(1-x/4)^-2$

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Find general formula for the terms


Deriving a formula to find the sum of a series.Find general formula for a seriesHow to find the general formula for this recursive problem?Demonstration of a simple formulaFinding general formula for sequenceGeneral formula for harmonic sequenceHow can I tell if the sequence re-cycles?The general term of such a recursion formulaFind an explicit formula for the recursive formulaThe general nth derivative formula of $(1-x/4)^-2$













5












$begingroup$



Find a general formula for the terms of the sequence



$$a_n=left frac117,frac10749,frac659343,frac48832401,frac3337116807,frac234569117649,dots right$$




I don't know how to approach this question as it is not arithmetic or geometric. I know the denominator is geometric increasing by a factor of $7$ but I can't find what the numerator should be for the general formula for the terms. Anyone know what it is?










share|cite|improve this question









New contributor




RaV1oLLi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 1




    $begingroup$
    Where does it come from?
    $endgroup$
    – Jean-Claude Arbaut
    Apr 21 at 17:14










  • $begingroup$
    My online calculus questions platform called Mobius(used to be Maple TA). I believe it must be a mistake as nothing I input is giving me the correct answer.
    $endgroup$
    – RaV1oLLi
    Apr 21 at 17:16










  • $begingroup$
    Have you put it into OEIS?
    $endgroup$
    – Dave
    Apr 21 at 17:22










  • $begingroup$
    This is in the OEIS, but not much else - oeis.org/A081657
    $endgroup$
    – Peter Foreman
    Apr 21 at 17:23






  • 1




    $begingroup$
    Nobody can tell you the correct answer. Whatever answer you give, the proposer can tell you that you are wrong. You an't win this kind of game.
    $endgroup$
    – Somos
    Apr 21 at 18:55















5












$begingroup$



Find a general formula for the terms of the sequence



$$a_n=left frac117,frac10749,frac659343,frac48832401,frac3337116807,frac234569117649,dots right$$




I don't know how to approach this question as it is not arithmetic or geometric. I know the denominator is geometric increasing by a factor of $7$ but I can't find what the numerator should be for the general formula for the terms. Anyone know what it is?










share|cite|improve this question









New contributor




RaV1oLLi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 1




    $begingroup$
    Where does it come from?
    $endgroup$
    – Jean-Claude Arbaut
    Apr 21 at 17:14










  • $begingroup$
    My online calculus questions platform called Mobius(used to be Maple TA). I believe it must be a mistake as nothing I input is giving me the correct answer.
    $endgroup$
    – RaV1oLLi
    Apr 21 at 17:16










  • $begingroup$
    Have you put it into OEIS?
    $endgroup$
    – Dave
    Apr 21 at 17:22










  • $begingroup$
    This is in the OEIS, but not much else - oeis.org/A081657
    $endgroup$
    – Peter Foreman
    Apr 21 at 17:23






  • 1




    $begingroup$
    Nobody can tell you the correct answer. Whatever answer you give, the proposer can tell you that you are wrong. You an't win this kind of game.
    $endgroup$
    – Somos
    Apr 21 at 18:55













5












5








5


2



$begingroup$



Find a general formula for the terms of the sequence



$$a_n=left frac117,frac10749,frac659343,frac48832401,frac3337116807,frac234569117649,dots right$$




I don't know how to approach this question as it is not arithmetic or geometric. I know the denominator is geometric increasing by a factor of $7$ but I can't find what the numerator should be for the general formula for the terms. Anyone know what it is?










share|cite|improve this question









New contributor




RaV1oLLi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$





Find a general formula for the terms of the sequence



$$a_n=left frac117,frac10749,frac659343,frac48832401,frac3337116807,frac234569117649,dots right$$




I don't know how to approach this question as it is not arithmetic or geometric. I know the denominator is geometric increasing by a factor of $7$ but I can't find what the numerator should be for the general formula for the terms. Anyone know what it is?







sequences-and-series






share|cite|improve this question









New contributor




RaV1oLLi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




RaV1oLLi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Apr 22 at 14:09









MarianD

2,3461619




2,3461619






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asked Apr 21 at 16:56









RaV1oLLiRaV1oLLi

292




292




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Check out our Code of Conduct.





New contributor





RaV1oLLi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






RaV1oLLi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 1




    $begingroup$
    Where does it come from?
    $endgroup$
    – Jean-Claude Arbaut
    Apr 21 at 17:14










  • $begingroup$
    My online calculus questions platform called Mobius(used to be Maple TA). I believe it must be a mistake as nothing I input is giving me the correct answer.
    $endgroup$
    – RaV1oLLi
    Apr 21 at 17:16










  • $begingroup$
    Have you put it into OEIS?
    $endgroup$
    – Dave
    Apr 21 at 17:22










  • $begingroup$
    This is in the OEIS, but not much else - oeis.org/A081657
    $endgroup$
    – Peter Foreman
    Apr 21 at 17:23






  • 1




    $begingroup$
    Nobody can tell you the correct answer. Whatever answer you give, the proposer can tell you that you are wrong. You an't win this kind of game.
    $endgroup$
    – Somos
    Apr 21 at 18:55












  • 1




    $begingroup$
    Where does it come from?
    $endgroup$
    – Jean-Claude Arbaut
    Apr 21 at 17:14










  • $begingroup$
    My online calculus questions platform called Mobius(used to be Maple TA). I believe it must be a mistake as nothing I input is giving me the correct answer.
    $endgroup$
    – RaV1oLLi
    Apr 21 at 17:16










  • $begingroup$
    Have you put it into OEIS?
    $endgroup$
    – Dave
    Apr 21 at 17:22










  • $begingroup$
    This is in the OEIS, but not much else - oeis.org/A081657
    $endgroup$
    – Peter Foreman
    Apr 21 at 17:23






  • 1




    $begingroup$
    Nobody can tell you the correct answer. Whatever answer you give, the proposer can tell you that you are wrong. You an't win this kind of game.
    $endgroup$
    – Somos
    Apr 21 at 18:55







1




1




$begingroup$
Where does it come from?
$endgroup$
– Jean-Claude Arbaut
Apr 21 at 17:14




$begingroup$
Where does it come from?
$endgroup$
– Jean-Claude Arbaut
Apr 21 at 17:14












$begingroup$
My online calculus questions platform called Mobius(used to be Maple TA). I believe it must be a mistake as nothing I input is giving me the correct answer.
$endgroup$
– RaV1oLLi
Apr 21 at 17:16




$begingroup$
My online calculus questions platform called Mobius(used to be Maple TA). I believe it must be a mistake as nothing I input is giving me the correct answer.
$endgroup$
– RaV1oLLi
Apr 21 at 17:16












$begingroup$
Have you put it into OEIS?
$endgroup$
– Dave
Apr 21 at 17:22




$begingroup$
Have you put it into OEIS?
$endgroup$
– Dave
Apr 21 at 17:22












$begingroup$
This is in the OEIS, but not much else - oeis.org/A081657
$endgroup$
– Peter Foreman
Apr 21 at 17:23




$begingroup$
This is in the OEIS, but not much else - oeis.org/A081657
$endgroup$
– Peter Foreman
Apr 21 at 17:23




1




1




$begingroup$
Nobody can tell you the correct answer. Whatever answer you give, the proposer can tell you that you are wrong. You an't win this kind of game.
$endgroup$
– Somos
Apr 21 at 18:55




$begingroup$
Nobody can tell you the correct answer. Whatever answer you give, the proposer can tell you that you are wrong. You an't win this kind of game.
$endgroup$
– Somos
Apr 21 at 18:55










3 Answers
3






active

oldest

votes


















6












$begingroup$

The general formula for an (infinite) sequence of (e. g. real) numbers from the finite number $n$ of its (first) members is in principle impossible, as the next (not listed) $(n+1)^mathrmth$ member may be an arbitrary number, and there is still a formula for expressing $a_1, dots, a_n, a_n+1,$ e. g. as a polynomial of order $n$:



$$a_k = sum_i=0^nb_ik^i,quad k = 1, dots,n+1$$



The process for finding coefficients $b_0, dots, b_n$ is straightforward enough.



In other words, if someone will find the formula for your "sequence", there is still the infinity number of other formulas, giving different sequences, but all of them producing your "sequence", i. e. $$frac117,frac10749,frac659343,frac48832401,frac3337116807,frac234569117649.$$




Note:



It means that all psychological tests of type




What is the next number of the sequence $1, 2, 3, 4, 5?$




are in principle meaningless ones, because you may tell "$1762$", and then show to surprised psychologist a formula supporting the correctness of your answer:



$$a_k = 439over30k^5-439over 2k^4+7463over 6k^3-6585over 2k^2+60158over 15k-1756$$



If he/she will not trust you, launch SageMath, which will produce accurate, non-rounded results, and write commands



sage: var("k")
sage: a(k) = (439/30)*k^5 - (439/2)*k^4+(7463/6)*k^3-(6585/2)*k^2+(60158/15)*k-1756
sage: a(1), a(2), a(3), a(4), a(5), a(6), a(7), a(8)


to obtain the result




(1, 2, 3, 4, 5, 1762, 10543, 36884)



(and to give the psychologist two more members for free).




Note 2:



It doesn't mean that there is not a simpler formula - including a recurrent one or other "recipe" - for the same (finite) sequence.



For example, there is so simple one for the rather not so trivial sequence



$$colorblue1, 11, 21, 1211, 111221$$



that even 6-7 year-old child is able to write down the next element ($colorred312211$) if you tell it the rule, or - perhaps - even without telling it.



No, you have no chance to discover this simple rule (supposing your age is $10$+). Don't waste your time. It's a good advice, believe me.



(Googling for it is a much better approach.)






share|cite|improve this answer











$endgroup$




















    5












    $begingroup$

    The general term seems to be
    $$a_n=frac2(7^n)+(-3)^n7^n=2+left(-frac37right)^n$$
    But the last term is given by
    $$a_6=2-left(frac37right)^6$$
    so this formula does not always work. A suitable formula could be
    $$a_n=begincases2-left(frac37right)^n&nequiv0mod6\
    2+left(-frac37right)^n &textotherwise
    endcases$$






    share|cite|improve this answer











    $endgroup$




















      0












      $begingroup$

      The numerator of the $n$-th term seems to be
      $$2cdot 7^n+(-1)^n 3^n. $$






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        May I ask how you got to this answer? Or is trial and error the only method?Edit: I inputted this answer into the online question and it came out as incorrect.
        $endgroup$
        – RaV1oLLi
        Apr 21 at 17:07











      • $begingroup$
        I was thinking the same, but what about the last term? $234569ne 2(7^6)+3^6$
        $endgroup$
        – Peter Foreman
        Apr 21 at 17:07







      • 1




        $begingroup$
        @PeterForeman Arg, yes, the last is $2(7)^6colorred-3^6$!
        $endgroup$
        – Jean-Claude Arbaut
        Apr 21 at 17:10







      • 1




        $begingroup$
        @RaV1oLLi: I noticed the numerators are the successive powers of $7$, and the numerators were not far from twice the denominator, with a corrective term. So I subtracted twice the denominators from the numerators, and tried to see a pattern. I tested only the 5 first terms.
        $endgroup$
        – Bernard
        Apr 21 at 17:20












      Your Answer








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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      The general formula for an (infinite) sequence of (e. g. real) numbers from the finite number $n$ of its (first) members is in principle impossible, as the next (not listed) $(n+1)^mathrmth$ member may be an arbitrary number, and there is still a formula for expressing $a_1, dots, a_n, a_n+1,$ e. g. as a polynomial of order $n$:



      $$a_k = sum_i=0^nb_ik^i,quad k = 1, dots,n+1$$



      The process for finding coefficients $b_0, dots, b_n$ is straightforward enough.



      In other words, if someone will find the formula for your "sequence", there is still the infinity number of other formulas, giving different sequences, but all of them producing your "sequence", i. e. $$frac117,frac10749,frac659343,frac48832401,frac3337116807,frac234569117649.$$




      Note:



      It means that all psychological tests of type




      What is the next number of the sequence $1, 2, 3, 4, 5?$




      are in principle meaningless ones, because you may tell "$1762$", and then show to surprised psychologist a formula supporting the correctness of your answer:



      $$a_k = 439over30k^5-439over 2k^4+7463over 6k^3-6585over 2k^2+60158over 15k-1756$$



      If he/she will not trust you, launch SageMath, which will produce accurate, non-rounded results, and write commands



      sage: var("k")
      sage: a(k) = (439/30)*k^5 - (439/2)*k^4+(7463/6)*k^3-(6585/2)*k^2+(60158/15)*k-1756
      sage: a(1), a(2), a(3), a(4), a(5), a(6), a(7), a(8)


      to obtain the result




      (1, 2, 3, 4, 5, 1762, 10543, 36884)



      (and to give the psychologist two more members for free).




      Note 2:



      It doesn't mean that there is not a simpler formula - including a recurrent one or other "recipe" - for the same (finite) sequence.



      For example, there is so simple one for the rather not so trivial sequence



      $$colorblue1, 11, 21, 1211, 111221$$



      that even 6-7 year-old child is able to write down the next element ($colorred312211$) if you tell it the rule, or - perhaps - even without telling it.



      No, you have no chance to discover this simple rule (supposing your age is $10$+). Don't waste your time. It's a good advice, believe me.



      (Googling for it is a much better approach.)






      share|cite|improve this answer











      $endgroup$

















        6












        $begingroup$

        The general formula for an (infinite) sequence of (e. g. real) numbers from the finite number $n$ of its (first) members is in principle impossible, as the next (not listed) $(n+1)^mathrmth$ member may be an arbitrary number, and there is still a formula for expressing $a_1, dots, a_n, a_n+1,$ e. g. as a polynomial of order $n$:



        $$a_k = sum_i=0^nb_ik^i,quad k = 1, dots,n+1$$



        The process for finding coefficients $b_0, dots, b_n$ is straightforward enough.



        In other words, if someone will find the formula for your "sequence", there is still the infinity number of other formulas, giving different sequences, but all of them producing your "sequence", i. e. $$frac117,frac10749,frac659343,frac48832401,frac3337116807,frac234569117649.$$




        Note:



        It means that all psychological tests of type




        What is the next number of the sequence $1, 2, 3, 4, 5?$




        are in principle meaningless ones, because you may tell "$1762$", and then show to surprised psychologist a formula supporting the correctness of your answer:



        $$a_k = 439over30k^5-439over 2k^4+7463over 6k^3-6585over 2k^2+60158over 15k-1756$$



        If he/she will not trust you, launch SageMath, which will produce accurate, non-rounded results, and write commands



        sage: var("k")
        sage: a(k) = (439/30)*k^5 - (439/2)*k^4+(7463/6)*k^3-(6585/2)*k^2+(60158/15)*k-1756
        sage: a(1), a(2), a(3), a(4), a(5), a(6), a(7), a(8)


        to obtain the result




        (1, 2, 3, 4, 5, 1762, 10543, 36884)



        (and to give the psychologist two more members for free).




        Note 2:



        It doesn't mean that there is not a simpler formula - including a recurrent one or other "recipe" - for the same (finite) sequence.



        For example, there is so simple one for the rather not so trivial sequence



        $$colorblue1, 11, 21, 1211, 111221$$



        that even 6-7 year-old child is able to write down the next element ($colorred312211$) if you tell it the rule, or - perhaps - even without telling it.



        No, you have no chance to discover this simple rule (supposing your age is $10$+). Don't waste your time. It's a good advice, believe me.



        (Googling for it is a much better approach.)






        share|cite|improve this answer











        $endgroup$















          6












          6








          6





          $begingroup$

          The general formula for an (infinite) sequence of (e. g. real) numbers from the finite number $n$ of its (first) members is in principle impossible, as the next (not listed) $(n+1)^mathrmth$ member may be an arbitrary number, and there is still a formula for expressing $a_1, dots, a_n, a_n+1,$ e. g. as a polynomial of order $n$:



          $$a_k = sum_i=0^nb_ik^i,quad k = 1, dots,n+1$$



          The process for finding coefficients $b_0, dots, b_n$ is straightforward enough.



          In other words, if someone will find the formula for your "sequence", there is still the infinity number of other formulas, giving different sequences, but all of them producing your "sequence", i. e. $$frac117,frac10749,frac659343,frac48832401,frac3337116807,frac234569117649.$$




          Note:



          It means that all psychological tests of type




          What is the next number of the sequence $1, 2, 3, 4, 5?$




          are in principle meaningless ones, because you may tell "$1762$", and then show to surprised psychologist a formula supporting the correctness of your answer:



          $$a_k = 439over30k^5-439over 2k^4+7463over 6k^3-6585over 2k^2+60158over 15k-1756$$



          If he/she will not trust you, launch SageMath, which will produce accurate, non-rounded results, and write commands



          sage: var("k")
          sage: a(k) = (439/30)*k^5 - (439/2)*k^4+(7463/6)*k^3-(6585/2)*k^2+(60158/15)*k-1756
          sage: a(1), a(2), a(3), a(4), a(5), a(6), a(7), a(8)


          to obtain the result




          (1, 2, 3, 4, 5, 1762, 10543, 36884)



          (and to give the psychologist two more members for free).




          Note 2:



          It doesn't mean that there is not a simpler formula - including a recurrent one or other "recipe" - for the same (finite) sequence.



          For example, there is so simple one for the rather not so trivial sequence



          $$colorblue1, 11, 21, 1211, 111221$$



          that even 6-7 year-old child is able to write down the next element ($colorred312211$) if you tell it the rule, or - perhaps - even without telling it.



          No, you have no chance to discover this simple rule (supposing your age is $10$+). Don't waste your time. It's a good advice, believe me.



          (Googling for it is a much better approach.)






          share|cite|improve this answer











          $endgroup$



          The general formula for an (infinite) sequence of (e. g. real) numbers from the finite number $n$ of its (first) members is in principle impossible, as the next (not listed) $(n+1)^mathrmth$ member may be an arbitrary number, and there is still a formula for expressing $a_1, dots, a_n, a_n+1,$ e. g. as a polynomial of order $n$:



          $$a_k = sum_i=0^nb_ik^i,quad k = 1, dots,n+1$$



          The process for finding coefficients $b_0, dots, b_n$ is straightforward enough.



          In other words, if someone will find the formula for your "sequence", there is still the infinity number of other formulas, giving different sequences, but all of them producing your "sequence", i. e. $$frac117,frac10749,frac659343,frac48832401,frac3337116807,frac234569117649.$$




          Note:



          It means that all psychological tests of type




          What is the next number of the sequence $1, 2, 3, 4, 5?$




          are in principle meaningless ones, because you may tell "$1762$", and then show to surprised psychologist a formula supporting the correctness of your answer:



          $$a_k = 439over30k^5-439over 2k^4+7463over 6k^3-6585over 2k^2+60158over 15k-1756$$



          If he/she will not trust you, launch SageMath, which will produce accurate, non-rounded results, and write commands



          sage: var("k")
          sage: a(k) = (439/30)*k^5 - (439/2)*k^4+(7463/6)*k^3-(6585/2)*k^2+(60158/15)*k-1756
          sage: a(1), a(2), a(3), a(4), a(5), a(6), a(7), a(8)


          to obtain the result




          (1, 2, 3, 4, 5, 1762, 10543, 36884)



          (and to give the psychologist two more members for free).




          Note 2:



          It doesn't mean that there is not a simpler formula - including a recurrent one or other "recipe" - for the same (finite) sequence.



          For example, there is so simple one for the rather not so trivial sequence



          $$colorblue1, 11, 21, 1211, 111221$$



          that even 6-7 year-old child is able to write down the next element ($colorred312211$) if you tell it the rule, or - perhaps - even without telling it.



          No, you have no chance to discover this simple rule (supposing your age is $10$+). Don't waste your time. It's a good advice, believe me.



          (Googling for it is a much better approach.)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 22 at 3:08

























          answered Apr 21 at 17:34









          MarianDMarianD

          2,3461619




          2,3461619





















              5












              $begingroup$

              The general term seems to be
              $$a_n=frac2(7^n)+(-3)^n7^n=2+left(-frac37right)^n$$
              But the last term is given by
              $$a_6=2-left(frac37right)^6$$
              so this formula does not always work. A suitable formula could be
              $$a_n=begincases2-left(frac37right)^n&nequiv0mod6\
              2+left(-frac37right)^n &textotherwise
              endcases$$






              share|cite|improve this answer











              $endgroup$

















                5












                $begingroup$

                The general term seems to be
                $$a_n=frac2(7^n)+(-3)^n7^n=2+left(-frac37right)^n$$
                But the last term is given by
                $$a_6=2-left(frac37right)^6$$
                so this formula does not always work. A suitable formula could be
                $$a_n=begincases2-left(frac37right)^n&nequiv0mod6\
                2+left(-frac37right)^n &textotherwise
                endcases$$






                share|cite|improve this answer











                $endgroup$















                  5












                  5








                  5





                  $begingroup$

                  The general term seems to be
                  $$a_n=frac2(7^n)+(-3)^n7^n=2+left(-frac37right)^n$$
                  But the last term is given by
                  $$a_6=2-left(frac37right)^6$$
                  so this formula does not always work. A suitable formula could be
                  $$a_n=begincases2-left(frac37right)^n&nequiv0mod6\
                  2+left(-frac37right)^n &textotherwise
                  endcases$$






                  share|cite|improve this answer











                  $endgroup$



                  The general term seems to be
                  $$a_n=frac2(7^n)+(-3)^n7^n=2+left(-frac37right)^n$$
                  But the last term is given by
                  $$a_6=2-left(frac37right)^6$$
                  so this formula does not always work. A suitable formula could be
                  $$a_n=begincases2-left(frac37right)^n&nequiv0mod6\
                  2+left(-frac37right)^n &textotherwise
                  endcases$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Apr 21 at 17:18

























                  answered Apr 21 at 17:13









                  Peter ForemanPeter Foreman

                  8,7701321




                  8,7701321





















                      0












                      $begingroup$

                      The numerator of the $n$-th term seems to be
                      $$2cdot 7^n+(-1)^n 3^n. $$






                      share|cite|improve this answer









                      $endgroup$












                      • $begingroup$
                        May I ask how you got to this answer? Or is trial and error the only method?Edit: I inputted this answer into the online question and it came out as incorrect.
                        $endgroup$
                        – RaV1oLLi
                        Apr 21 at 17:07











                      • $begingroup$
                        I was thinking the same, but what about the last term? $234569ne 2(7^6)+3^6$
                        $endgroup$
                        – Peter Foreman
                        Apr 21 at 17:07







                      • 1




                        $begingroup$
                        @PeterForeman Arg, yes, the last is $2(7)^6colorred-3^6$!
                        $endgroup$
                        – Jean-Claude Arbaut
                        Apr 21 at 17:10







                      • 1




                        $begingroup$
                        @RaV1oLLi: I noticed the numerators are the successive powers of $7$, and the numerators were not far from twice the denominator, with a corrective term. So I subtracted twice the denominators from the numerators, and tried to see a pattern. I tested only the 5 first terms.
                        $endgroup$
                        – Bernard
                        Apr 21 at 17:20
















                      0












                      $begingroup$

                      The numerator of the $n$-th term seems to be
                      $$2cdot 7^n+(-1)^n 3^n. $$






                      share|cite|improve this answer









                      $endgroup$












                      • $begingroup$
                        May I ask how you got to this answer? Or is trial and error the only method?Edit: I inputted this answer into the online question and it came out as incorrect.
                        $endgroup$
                        – RaV1oLLi
                        Apr 21 at 17:07











                      • $begingroup$
                        I was thinking the same, but what about the last term? $234569ne 2(7^6)+3^6$
                        $endgroup$
                        – Peter Foreman
                        Apr 21 at 17:07







                      • 1




                        $begingroup$
                        @PeterForeman Arg, yes, the last is $2(7)^6colorred-3^6$!
                        $endgroup$
                        – Jean-Claude Arbaut
                        Apr 21 at 17:10







                      • 1




                        $begingroup$
                        @RaV1oLLi: I noticed the numerators are the successive powers of $7$, and the numerators were not far from twice the denominator, with a corrective term. So I subtracted twice the denominators from the numerators, and tried to see a pattern. I tested only the 5 first terms.
                        $endgroup$
                        – Bernard
                        Apr 21 at 17:20














                      0












                      0








                      0





                      $begingroup$

                      The numerator of the $n$-th term seems to be
                      $$2cdot 7^n+(-1)^n 3^n. $$






                      share|cite|improve this answer









                      $endgroup$



                      The numerator of the $n$-th term seems to be
                      $$2cdot 7^n+(-1)^n 3^n. $$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Apr 21 at 17:05









                      BernardBernard

                      125k743119




                      125k743119











                      • $begingroup$
                        May I ask how you got to this answer? Or is trial and error the only method?Edit: I inputted this answer into the online question and it came out as incorrect.
                        $endgroup$
                        – RaV1oLLi
                        Apr 21 at 17:07











                      • $begingroup$
                        I was thinking the same, but what about the last term? $234569ne 2(7^6)+3^6$
                        $endgroup$
                        – Peter Foreman
                        Apr 21 at 17:07







                      • 1




                        $begingroup$
                        @PeterForeman Arg, yes, the last is $2(7)^6colorred-3^6$!
                        $endgroup$
                        – Jean-Claude Arbaut
                        Apr 21 at 17:10







                      • 1




                        $begingroup$
                        @RaV1oLLi: I noticed the numerators are the successive powers of $7$, and the numerators were not far from twice the denominator, with a corrective term. So I subtracted twice the denominators from the numerators, and tried to see a pattern. I tested only the 5 first terms.
                        $endgroup$
                        – Bernard
                        Apr 21 at 17:20

















                      • $begingroup$
                        May I ask how you got to this answer? Or is trial and error the only method?Edit: I inputted this answer into the online question and it came out as incorrect.
                        $endgroup$
                        – RaV1oLLi
                        Apr 21 at 17:07











                      • $begingroup$
                        I was thinking the same, but what about the last term? $234569ne 2(7^6)+3^6$
                        $endgroup$
                        – Peter Foreman
                        Apr 21 at 17:07







                      • 1




                        $begingroup$
                        @PeterForeman Arg, yes, the last is $2(7)^6colorred-3^6$!
                        $endgroup$
                        – Jean-Claude Arbaut
                        Apr 21 at 17:10







                      • 1




                        $begingroup$
                        @RaV1oLLi: I noticed the numerators are the successive powers of $7$, and the numerators were not far from twice the denominator, with a corrective term. So I subtracted twice the denominators from the numerators, and tried to see a pattern. I tested only the 5 first terms.
                        $endgroup$
                        – Bernard
                        Apr 21 at 17:20
















                      $begingroup$
                      May I ask how you got to this answer? Or is trial and error the only method?Edit: I inputted this answer into the online question and it came out as incorrect.
                      $endgroup$
                      – RaV1oLLi
                      Apr 21 at 17:07





                      $begingroup$
                      May I ask how you got to this answer? Or is trial and error the only method?Edit: I inputted this answer into the online question and it came out as incorrect.
                      $endgroup$
                      – RaV1oLLi
                      Apr 21 at 17:07













                      $begingroup$
                      I was thinking the same, but what about the last term? $234569ne 2(7^6)+3^6$
                      $endgroup$
                      – Peter Foreman
                      Apr 21 at 17:07





                      $begingroup$
                      I was thinking the same, but what about the last term? $234569ne 2(7^6)+3^6$
                      $endgroup$
                      – Peter Foreman
                      Apr 21 at 17:07





                      1




                      1




                      $begingroup$
                      @PeterForeman Arg, yes, the last is $2(7)^6colorred-3^6$!
                      $endgroup$
                      – Jean-Claude Arbaut
                      Apr 21 at 17:10





                      $begingroup$
                      @PeterForeman Arg, yes, the last is $2(7)^6colorred-3^6$!
                      $endgroup$
                      – Jean-Claude Arbaut
                      Apr 21 at 17:10





                      1




                      1




                      $begingroup$
                      @RaV1oLLi: I noticed the numerators are the successive powers of $7$, and the numerators were not far from twice the denominator, with a corrective term. So I subtracted twice the denominators from the numerators, and tried to see a pattern. I tested only the 5 first terms.
                      $endgroup$
                      – Bernard
                      Apr 21 at 17:20





                      $begingroup$
                      @RaV1oLLi: I noticed the numerators are the successive powers of $7$, and the numerators were not far from twice the denominator, with a corrective term. So I subtracted twice the denominators from the numerators, and tried to see a pattern. I tested only the 5 first terms.
                      $endgroup$
                      – Bernard
                      Apr 21 at 17:20











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