How do I describe the progression $c+ba^{-(n-1)}$? I'm seeking an efficient sum to arbitrary $n$.












2












$begingroup$


How do I characterise a progression of the form:
$$
c+ba^{-(n-1)}
$$



where integer $n > 0$.



My motivation is that I'm trying to find an efficient sum for this progression, to an arbitrary $n$. (The existing function sums by iterative addition, which is very inefficient for even moderately large $n$.)



From asking earlier I am informed this is not a geometric progression (and hence I can't use the sum formula for a geometric progression).



So, what kind of progression is this, and how do I efficiently sum it, ideally with arithmetic no more complex than exponentiation?






sequences-and-series algorithms







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @dmtri, thanks you were faster than my edit :-) The question now uses $n$ in the expression.
    $endgroup$
    – bignose
    Jan 5 at 6:20






  • 1




    $begingroup$
    Note that $$displaystylesum_1^nBig[c+frac b{a^{k-1}}Big]=sum_1^n c+sum_1^nfrac b{a^{k-1}}=nc+bBig[frac{1-a^{-n}}{1-a^{-1}}Big]$$ where the second term is the sum of a geometric progression.
    $endgroup$
    – Shubham Johri
    Jan 5 at 6:25












  • $begingroup$
    @ShubhamJohri, would you like to put that into an answer? If not, I can do it.
    $endgroup$
    – bignose
    Jan 6 at 1:22












  • $begingroup$
    Sure, there you go
    $endgroup$
    – Shubham Johri
    Jan 6 at 6:23


















2












$begingroup$


How do I characterise a progression of the form:
$$
c+ba^{-(n-1)}
$$



where integer $n > 0$.



My motivation is that I'm trying to find an efficient sum for this progression, to an arbitrary $n$. (The existing function sums by iterative addition, which is very inefficient for even moderately large $n$.)



From asking earlier I am informed this is not a geometric progression (and hence I can't use the sum formula for a geometric progression).



So, what kind of progression is this, and how do I efficiently sum it, ideally with arithmetic no more complex than exponentiation?






sequences-and-series algorithms







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @dmtri, thanks you were faster than my edit :-) The question now uses $n$ in the expression.
    $endgroup$
    – bignose
    Jan 5 at 6:20






  • 1




    $begingroup$
    Note that $$displaystylesum_1^nBig[c+frac b{a^{k-1}}Big]=sum_1^n c+sum_1^nfrac b{a^{k-1}}=nc+bBig[frac{1-a^{-n}}{1-a^{-1}}Big]$$ where the second term is the sum of a geometric progression.
    $endgroup$
    – Shubham Johri
    Jan 5 at 6:25












  • $begingroup$
    @ShubhamJohri, would you like to put that into an answer? If not, I can do it.
    $endgroup$
    – bignose
    Jan 6 at 1:22












  • $begingroup$
    Sure, there you go
    $endgroup$
    – Shubham Johri
    Jan 6 at 6:23
















2












2








2





$begingroup$


How do I characterise a progression of the form:
$$
c+ba^{-(n-1)}
$$



where integer $n > 0$.



My motivation is that I'm trying to find an efficient sum for this progression, to an arbitrary $n$. (The existing function sums by iterative addition, which is very inefficient for even moderately large $n$.)



From asking earlier I am informed this is not a geometric progression (and hence I can't use the sum formula for a geometric progression).



So, what kind of progression is this, and how do I efficiently sum it, ideally with arithmetic no more complex than exponentiation?






sequences-and-series algorithms







share|cite|improve this question











$endgroup$




How do I characterise a progression of the form:
$$
c+ba^{-(n-1)}
$$



where integer $n > 0$.



My motivation is that I'm trying to find an efficient sum for this progression, to an arbitrary $n$. (The existing function sums by iterative addition, which is very inefficient for even moderately large $n$.)



From asking earlier I am informed this is not a geometric progression (and hence I can't use the sum formula for a geometric progression).



So, what kind of progression is this, and how do I efficiently sum it, ideally with arithmetic no more complex than exponentiation?






sequences-and-series algorithms




sequences-and-series algorithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 6:26









Blue

47.7k870151




47.7k870151










asked Jan 5 at 6:15









bignosebignose

1255




1255








  • 1




    $begingroup$
    @dmtri, thanks you were faster than my edit :-) The question now uses $n$ in the expression.
    $endgroup$
    – bignose
    Jan 5 at 6:20






  • 1




    $begingroup$
    Note that $$displaystylesum_1^nBig[c+frac b{a^{k-1}}Big]=sum_1^n c+sum_1^nfrac b{a^{k-1}}=nc+bBig[frac{1-a^{-n}}{1-a^{-1}}Big]$$ where the second term is the sum of a geometric progression.
    $endgroup$
    – Shubham Johri
    Jan 5 at 6:25












  • $begingroup$
    @ShubhamJohri, would you like to put that into an answer? If not, I can do it.
    $endgroup$
    – bignose
    Jan 6 at 1:22












  • $begingroup$
    Sure, there you go
    $endgroup$
    – Shubham Johri
    Jan 6 at 6:23
















  • 1




    $begingroup$
    @dmtri, thanks you were faster than my edit :-) The question now uses $n$ in the expression.
    $endgroup$
    – bignose
    Jan 5 at 6:20






  • 1




    $begingroup$
    Note that $$displaystylesum_1^nBig[c+frac b{a^{k-1}}Big]=sum_1^n c+sum_1^nfrac b{a^{k-1}}=nc+bBig[frac{1-a^{-n}}{1-a^{-1}}Big]$$ where the second term is the sum of a geometric progression.
    $endgroup$
    – Shubham Johri
    Jan 5 at 6:25












  • $begingroup$
    @ShubhamJohri, would you like to put that into an answer? If not, I can do it.
    $endgroup$
    – bignose
    Jan 6 at 1:22












  • $begingroup$
    Sure, there you go
    $endgroup$
    – Shubham Johri
    Jan 6 at 6:23










1




1




$begingroup$
@dmtri, thanks you were faster than my edit :-) The question now uses $n$ in the expression.
$endgroup$
– bignose
Jan 5 at 6:20




$begingroup$
@dmtri, thanks you were faster than my edit :-) The question now uses $n$ in the expression.
$endgroup$
– bignose
Jan 5 at 6:20




1




1




$begingroup$
Note that $$displaystylesum_1^nBig[c+frac b{a^{k-1}}Big]=sum_1^n c+sum_1^nfrac b{a^{k-1}}=nc+bBig[frac{1-a^{-n}}{1-a^{-1}}Big]$$ where the second term is the sum of a geometric progression.
$endgroup$
– Shubham Johri
Jan 5 at 6:25






$begingroup$
Note that $$displaystylesum_1^nBig[c+frac b{a^{k-1}}Big]=sum_1^n c+sum_1^nfrac b{a^{k-1}}=nc+bBig[frac{1-a^{-n}}{1-a^{-1}}Big]$$ where the second term is the sum of a geometric progression.
$endgroup$
– Shubham Johri
Jan 5 at 6:25














$begingroup$
@ShubhamJohri, would you like to put that into an answer? If not, I can do it.
$endgroup$
– bignose
Jan 6 at 1:22






$begingroup$
@ShubhamJohri, would you like to put that into an answer? If not, I can do it.
$endgroup$
– bignose
Jan 6 at 1:22














$begingroup$
Sure, there you go
$endgroup$
– Shubham Johri
Jan 6 at 6:23






$begingroup$
Sure, there you go
$endgroup$
– Shubham Johri
Jan 6 at 6:23












2 Answers
2






active

oldest

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3












$begingroup$

You have
$$u_n = c + frac{b}{a^{n-1}}$$ and you want to find
$$S_n = sum_{k=1}^{n}u_k = sum_{k=1}^{n}left( c + frac{b}{a^{k-1}}right) =cn+bleft(1+frac{1}{a^2}+cdots +frac{1}{a^{n-1}}right) $$
$$ = cn+frac{b}{a^{n-1}}left(frac{a^n-1}{a-1}right).$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Note that $$sum_1^nBig[c+frac b{a^{k-1}}Big]=sum_1^n c+sum_1^nfrac b{a^{k-1}}=nc+bBig[frac{1-a^{-n}}{1-a^{-1}}Big]$$where the second term is the sum of a geometric progression.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      You have
      $$u_n = c + frac{b}{a^{n-1}}$$ and you want to find
      $$S_n = sum_{k=1}^{n}u_k = sum_{k=1}^{n}left( c + frac{b}{a^{k-1}}right) =cn+bleft(1+frac{1}{a^2}+cdots +frac{1}{a^{n-1}}right) $$
      $$ = cn+frac{b}{a^{n-1}}left(frac{a^n-1}{a-1}right).$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        You have
        $$u_n = c + frac{b}{a^{n-1}}$$ and you want to find
        $$S_n = sum_{k=1}^{n}u_k = sum_{k=1}^{n}left( c + frac{b}{a^{k-1}}right) =cn+bleft(1+frac{1}{a^2}+cdots +frac{1}{a^{n-1}}right) $$
        $$ = cn+frac{b}{a^{n-1}}left(frac{a^n-1}{a-1}right).$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          You have
          $$u_n = c + frac{b}{a^{n-1}}$$ and you want to find
          $$S_n = sum_{k=1}^{n}u_k = sum_{k=1}^{n}left( c + frac{b}{a^{k-1}}right) =cn+bleft(1+frac{1}{a^2}+cdots +frac{1}{a^{n-1}}right) $$
          $$ = cn+frac{b}{a^{n-1}}left(frac{a^n-1}{a-1}right).$$






          share|cite|improve this answer









          $endgroup$



          You have
          $$u_n = c + frac{b}{a^{n-1}}$$ and you want to find
          $$S_n = sum_{k=1}^{n}u_k = sum_{k=1}^{n}left( c + frac{b}{a^{k-1}}right) =cn+bleft(1+frac{1}{a^2}+cdots +frac{1}{a^{n-1}}right) $$
          $$ = cn+frac{b}{a^{n-1}}left(frac{a^n-1}{a-1}right).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 5 at 6:26









          Hello_WorldHello_World

          4,15521630




          4,15521630























              0












              $begingroup$

              Note that $$sum_1^nBig[c+frac b{a^{k-1}}Big]=sum_1^n c+sum_1^nfrac b{a^{k-1}}=nc+bBig[frac{1-a^{-n}}{1-a^{-1}}Big]$$where the second term is the sum of a geometric progression.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Note that $$sum_1^nBig[c+frac b{a^{k-1}}Big]=sum_1^n c+sum_1^nfrac b{a^{k-1}}=nc+bBig[frac{1-a^{-n}}{1-a^{-1}}Big]$$where the second term is the sum of a geometric progression.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Note that $$sum_1^nBig[c+frac b{a^{k-1}}Big]=sum_1^n c+sum_1^nfrac b{a^{k-1}}=nc+bBig[frac{1-a^{-n}}{1-a^{-1}}Big]$$where the second term is the sum of a geometric progression.






                  share|cite|improve this answer









                  $endgroup$



                  Note that $$sum_1^nBig[c+frac b{a^{k-1}}Big]=sum_1^n c+sum_1^nfrac b{a^{k-1}}=nc+bBig[frac{1-a^{-n}}{1-a^{-1}}Big]$$where the second term is the sum of a geometric progression.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 6:21









                  Shubham JohriShubham Johri

                  4,666717




                  4,666717






























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