Geometric series starting from -n












1














How do I find the following series:



$$sum^n_{k = -n}ar^k$$



I'm not sure how to manipulate the standard geometric series to get the desired result.



Any direction would be appreciated.






sequences-and-series







share|cite|improve this question




















  • 2




    Hint: $sumlimits_{k=-n}^n f(k) = left(sumlimits_{k=-n}^{-1}f(k)right)+left(sumlimits_{k=0}^nf(k)right)$ and $sumlimits_{k=-n}^{-1}f(k)=sumlimits_{k=1}^n f(-k)$
    – JMoravitz
    Nov 20 at 1:58






  • 2




    Alternatively $sumlimits_{k=-n}^n f(k) = sumlimits_{k=0}^{2n} f(k-n)$
    – JMoravitz
    Nov 20 at 2:00
















1














How do I find the following series:



$$sum^n_{k = -n}ar^k$$



I'm not sure how to manipulate the standard geometric series to get the desired result.



Any direction would be appreciated.






sequences-and-series







share|cite|improve this question




















  • 2




    Hint: $sumlimits_{k=-n}^n f(k) = left(sumlimits_{k=-n}^{-1}f(k)right)+left(sumlimits_{k=0}^nf(k)right)$ and $sumlimits_{k=-n}^{-1}f(k)=sumlimits_{k=1}^n f(-k)$
    – JMoravitz
    Nov 20 at 1:58






  • 2




    Alternatively $sumlimits_{k=-n}^n f(k) = sumlimits_{k=0}^{2n} f(k-n)$
    – JMoravitz
    Nov 20 at 2:00














1












1








1







How do I find the following series:



$$sum^n_{k = -n}ar^k$$



I'm not sure how to manipulate the standard geometric series to get the desired result.



Any direction would be appreciated.






sequences-and-series







share|cite|improve this question















How do I find the following series:



$$sum^n_{k = -n}ar^k$$



I'm not sure how to manipulate the standard geometric series to get the desired result.



Any direction would be appreciated.






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 at 1:59









amWhy

191k28224439




191k28224439










asked Nov 20 at 1:57









51n84d

383




383








  • 2




    Hint: $sumlimits_{k=-n}^n f(k) = left(sumlimits_{k=-n}^{-1}f(k)right)+left(sumlimits_{k=0}^nf(k)right)$ and $sumlimits_{k=-n}^{-1}f(k)=sumlimits_{k=1}^n f(-k)$
    – JMoravitz
    Nov 20 at 1:58






  • 2




    Alternatively $sumlimits_{k=-n}^n f(k) = sumlimits_{k=0}^{2n} f(k-n)$
    – JMoravitz
    Nov 20 at 2:00














  • 2




    Hint: $sumlimits_{k=-n}^n f(k) = left(sumlimits_{k=-n}^{-1}f(k)right)+left(sumlimits_{k=0}^nf(k)right)$ and $sumlimits_{k=-n}^{-1}f(k)=sumlimits_{k=1}^n f(-k)$
    – JMoravitz
    Nov 20 at 1:58






  • 2




    Alternatively $sumlimits_{k=-n}^n f(k) = sumlimits_{k=0}^{2n} f(k-n)$
    – JMoravitz
    Nov 20 at 2:00








2




2




Hint: $sumlimits_{k=-n}^n f(k) = left(sumlimits_{k=-n}^{-1}f(k)right)+left(sumlimits_{k=0}^nf(k)right)$ and $sumlimits_{k=-n}^{-1}f(k)=sumlimits_{k=1}^n f(-k)$
– JMoravitz
Nov 20 at 1:58




Hint: $sumlimits_{k=-n}^n f(k) = left(sumlimits_{k=-n}^{-1}f(k)right)+left(sumlimits_{k=0}^nf(k)right)$ and $sumlimits_{k=-n}^{-1}f(k)=sumlimits_{k=1}^n f(-k)$
– JMoravitz
Nov 20 at 1:58




2




2




Alternatively $sumlimits_{k=-n}^n f(k) = sumlimits_{k=0}^{2n} f(k-n)$
– JMoravitz
Nov 20 at 2:00




Alternatively $sumlimits_{k=-n}^n f(k) = sumlimits_{k=0}^{2n} f(k-n)$
– JMoravitz
Nov 20 at 2:00










1 Answer
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1














You can write $$sum^n_{k = -n}ar^k=sum^n_{k = -n}ar^{k+n}r^{-n}=r^{-n}sum^{2n}_{m = 0}ar^m$$
by defining $m=k+n$






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    You can write $$sum^n_{k = -n}ar^k=sum^n_{k = -n}ar^{k+n}r^{-n}=r^{-n}sum^{2n}_{m = 0}ar^m$$
    by defining $m=k+n$






    share|cite|improve this answer


























      1














      You can write $$sum^n_{k = -n}ar^k=sum^n_{k = -n}ar^{k+n}r^{-n}=r^{-n}sum^{2n}_{m = 0}ar^m$$
      by defining $m=k+n$






      share|cite|improve this answer
























        1












        1








        1






        You can write $$sum^n_{k = -n}ar^k=sum^n_{k = -n}ar^{k+n}r^{-n}=r^{-n}sum^{2n}_{m = 0}ar^m$$
        by defining $m=k+n$






        share|cite|improve this answer












        You can write $$sum^n_{k = -n}ar^k=sum^n_{k = -n}ar^{k+n}r^{-n}=r^{-n}sum^{2n}_{m = 0}ar^m$$
        by defining $m=k+n$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 at 2:06









        Ross Millikan

        291k23196370




        291k23196370






























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