How to find the sum of $1+(1+r)s+(1+r+r^2)s^2+dots$?

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I was asked to find the geometric sum of the following:



$$1+(1+r)s+(1+r+r^2)s^2+dots$$



My first way to solve the problem is to expand the brackets, and sort them out into two different geometric series, and evaluate the separated series altogether:



$$1+(s+rs+dots)+(s^2+rs^2+r^2s^2+dots)$$



The only problem is it doesn't seem to work, as the third term, $(1+r+r^2 +r^3)s^3$ doesn't seem to fit the separated sequence correctly.



Any help would be appreciated.




sequences-and-series geometric-series




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  • 2




    Hint: $1+r+r^2 + cdots + r^k = frac1-r^k+11-r$
    – achille hui
    Aug 5 at 8:16














up vote
8
down vote

favorite
3












I was asked to find the geometric sum of the following:



$$1+(1+r)s+(1+r+r^2)s^2+dots$$



My first way to solve the problem is to expand the brackets, and sort them out into two different geometric series, and evaluate the separated series altogether:



$$1+(s+rs+dots)+(s^2+rs^2+r^2s^2+dots)$$



The only problem is it doesn't seem to work, as the third term, $(1+r+r^2 +r^3)s^3$ doesn't seem to fit the separated sequence correctly.



Any help would be appreciated.




sequences-and-series geometric-series




share|cite|improve this question

















  • 2




    Hint: $1+r+r^2 + cdots + r^k = frac1-r^k+11-r$
    – achille hui
    Aug 5 at 8:16












up vote
8
down vote

favorite
3









up vote
8
down vote

favorite
3






3





I was asked to find the geometric sum of the following:



$$1+(1+r)s+(1+r+r^2)s^2+dots$$



My first way to solve the problem is to expand the brackets, and sort them out into two different geometric series, and evaluate the separated series altogether:



$$1+(s+rs+dots)+(s^2+rs^2+r^2s^2+dots)$$



The only problem is it doesn't seem to work, as the third term, $(1+r+r^2 +r^3)s^3$ doesn't seem to fit the separated sequence correctly.



Any help would be appreciated.




sequences-and-series geometric-series




share|cite|improve this question













I was asked to find the geometric sum of the following:



$$1+(1+r)s+(1+r+r^2)s^2+dots$$



My first way to solve the problem is to expand the brackets, and sort them out into two different geometric series, and evaluate the separated series altogether:



$$1+(s+rs+dots)+(s^2+rs^2+r^2s^2+dots)$$



The only problem is it doesn't seem to work, as the third term, $(1+r+r^2 +r^3)s^3$ doesn't seem to fit the separated sequence correctly.



Any help would be appreciated.





sequences-and-series geometric-series





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 5 at 23:37









Asaf Karagila♦

292k31403733




292k31403733









asked Aug 5 at 8:11









Loo Soo Yong

694




694







  • 2




    Hint: $1+r+r^2 + cdots + r^k = frac1-r^k+11-r$
    – achille hui
    Aug 5 at 8:16












  • 2




    Hint: $1+r+r^2 + cdots + r^k = frac1-r^k+11-r$
    – achille hui
    Aug 5 at 8:16







2




2




Hint: $1+r+r^2 + cdots + r^k = frac1-r^k+11-r$
– achille hui
Aug 5 at 8:16




Hint: $1+r+r^2 + cdots + r^k = frac1-r^k+11-r$
– achille hui
Aug 5 at 8:16










3 Answers
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5
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accepted










You expanded the brackets, but did not actually group:




$$1+(1+r)s+(1+r+r^2)s^2+...=1+(s+rscolorred+...)+(s^2+rs^2+r^2s^2colorred+...).$$




Here is the way to group:
$$1+(1+r)s+(1+r+r^2)s^2+...=1+(colorreds+colorgreenrs)+(colorreds^2+colorgreenrs^2+colorbluer^2s^2)+cdots=\
(1+colorreds+colorreds^2+cdots)+(colorgreenrs+colorgreenrs^2+cdots)+(colorbluer^2s^2+r^2s^3+cdots)=\
frac11-s+fracrs1-s+fracr^2s^21-s+cdots=\
frac11-s(1+rs+r^2s^2+cdots)=cdots$$
Can you finish?






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    up vote
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    Your sum can be written as
    $$sum_n=0^inftyleft(sum_k=0^nr^kright)s^n=sum_n=0^inftyfrac1-r^n+11-rs^n=frac11-rsum_n=0^inftys^n-
    fracr1-rsum_n=0^infty(rs)^n.$$
    Can you take it from here?



    P.S. Here we are assuming that $|s|<1$, $|rs|<1$ and $rnot=1$. What happens when $r=1$?






    share|cite|improve this answer



















    • 1




      There's a typo in the last term (there should be a minus sign instead of a plus).
      – yoann
      Aug 5 at 10:10










    • @yoann Thanks for pointing out!!
      – Robert Z
      Aug 5 at 10:16

















    up vote
    7
    down vote













    If you multiply your series by $r-1$, you get$$(r-1)+(r^2-1)s+(r^3-1)s^2+cdots,tag1$$which is the sum of$$r+r^2s+r^3s^2+cdots$$with$$-1-s-s^2-cdots$$The sum of the first series is $frac r1-rs$, whereas the sum of the second one is $-frac11-s$. Therefore,$$(1)=frac1r-1left(frac r1-rs-frac11-sright)=frac1(1-s)(1-rs).$$






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    • 1




      @farruhota I've edited my answer. Thank you.
      – José Carlos Santos
      Aug 5 at 9:36










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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    5
    down vote



    accepted










    You expanded the brackets, but did not actually group:




    $$1+(1+r)s+(1+r+r^2)s^2+...=1+(s+rscolorred+...)+(s^2+rs^2+r^2s^2colorred+...).$$




    Here is the way to group:
    $$1+(1+r)s+(1+r+r^2)s^2+...=1+(colorreds+colorgreenrs)+(colorreds^2+colorgreenrs^2+colorbluer^2s^2)+cdots=\
    (1+colorreds+colorreds^2+cdots)+(colorgreenrs+colorgreenrs^2+cdots)+(colorbluer^2s^2+r^2s^3+cdots)=\
    frac11-s+fracrs1-s+fracr^2s^21-s+cdots=\
    frac11-s(1+rs+r^2s^2+cdots)=cdots$$
    Can you finish?






    share|cite|improve this answer

























      up vote
      5
      down vote



      accepted










      You expanded the brackets, but did not actually group:




      $$1+(1+r)s+(1+r+r^2)s^2+...=1+(s+rscolorred+...)+(s^2+rs^2+r^2s^2colorred+...).$$




      Here is the way to group:
      $$1+(1+r)s+(1+r+r^2)s^2+...=1+(colorreds+colorgreenrs)+(colorreds^2+colorgreenrs^2+colorbluer^2s^2)+cdots=\
      (1+colorreds+colorreds^2+cdots)+(colorgreenrs+colorgreenrs^2+cdots)+(colorbluer^2s^2+r^2s^3+cdots)=\
      frac11-s+fracrs1-s+fracr^2s^21-s+cdots=\
      frac11-s(1+rs+r^2s^2+cdots)=cdots$$
      Can you finish?






      share|cite|improve this answer























        up vote
        5
        down vote



        accepted







        up vote
        5
        down vote



        accepted






        You expanded the brackets, but did not actually group:




        $$1+(1+r)s+(1+r+r^2)s^2+...=1+(s+rscolorred+...)+(s^2+rs^2+r^2s^2colorred+...).$$




        Here is the way to group:
        $$1+(1+r)s+(1+r+r^2)s^2+...=1+(colorreds+colorgreenrs)+(colorreds^2+colorgreenrs^2+colorbluer^2s^2)+cdots=\
        (1+colorreds+colorreds^2+cdots)+(colorgreenrs+colorgreenrs^2+cdots)+(colorbluer^2s^2+r^2s^3+cdots)=\
        frac11-s+fracrs1-s+fracr^2s^21-s+cdots=\
        frac11-s(1+rs+r^2s^2+cdots)=cdots$$
        Can you finish?






        share|cite|improve this answer













        You expanded the brackets, but did not actually group:




        $$1+(1+r)s+(1+r+r^2)s^2+...=1+(s+rscolorred+...)+(s^2+rs^2+r^2s^2colorred+...).$$




        Here is the way to group:
        $$1+(1+r)s+(1+r+r^2)s^2+...=1+(colorreds+colorgreenrs)+(colorreds^2+colorgreenrs^2+colorbluer^2s^2)+cdots=\
        (1+colorreds+colorreds^2+cdots)+(colorgreenrs+colorgreenrs^2+cdots)+(colorbluer^2s^2+r^2s^3+cdots)=\
        frac11-s+fracrs1-s+fracr^2s^21-s+cdots=\
        frac11-s(1+rs+r^2s^2+cdots)=cdots$$
        Can you finish?







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 5 at 9:22









        farruhota

        13.8k2632




        13.8k2632




















            up vote
            15
            down vote













            Your sum can be written as
            $$sum_n=0^inftyleft(sum_k=0^nr^kright)s^n=sum_n=0^inftyfrac1-r^n+11-rs^n=frac11-rsum_n=0^inftys^n-
            fracr1-rsum_n=0^infty(rs)^n.$$
            Can you take it from here?



            P.S. Here we are assuming that $|s|<1$, $|rs|<1$ and $rnot=1$. What happens when $r=1$?






            share|cite|improve this answer



















            • 1




              There's a typo in the last term (there should be a minus sign instead of a plus).
              – yoann
              Aug 5 at 10:10










            • @yoann Thanks for pointing out!!
              – Robert Z
              Aug 5 at 10:16














            up vote
            15
            down vote













            Your sum can be written as
            $$sum_n=0^inftyleft(sum_k=0^nr^kright)s^n=sum_n=0^inftyfrac1-r^n+11-rs^n=frac11-rsum_n=0^inftys^n-
            fracr1-rsum_n=0^infty(rs)^n.$$
            Can you take it from here?



            P.S. Here we are assuming that $|s|<1$, $|rs|<1$ and $rnot=1$. What happens when $r=1$?






            share|cite|improve this answer



















            • 1




              There's a typo in the last term (there should be a minus sign instead of a plus).
              – yoann
              Aug 5 at 10:10










            • @yoann Thanks for pointing out!!
              – Robert Z
              Aug 5 at 10:16












            up vote
            15
            down vote










            up vote
            15
            down vote









            Your sum can be written as
            $$sum_n=0^inftyleft(sum_k=0^nr^kright)s^n=sum_n=0^inftyfrac1-r^n+11-rs^n=frac11-rsum_n=0^inftys^n-
            fracr1-rsum_n=0^infty(rs)^n.$$
            Can you take it from here?



            P.S. Here we are assuming that $|s|<1$, $|rs|<1$ and $rnot=1$. What happens when $r=1$?






            share|cite|improve this answer















            Your sum can be written as
            $$sum_n=0^inftyleft(sum_k=0^nr^kright)s^n=sum_n=0^inftyfrac1-r^n+11-rs^n=frac11-rsum_n=0^inftys^n-
            fracr1-rsum_n=0^infty(rs)^n.$$
            Can you take it from here?



            P.S. Here we are assuming that $|s|<1$, $|rs|<1$ and $rnot=1$. What happens when $r=1$?







            share|cite|improve this answer















            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 5 at 10:15


























            answered Aug 5 at 8:17









            Robert Z

            84.2k955123




            84.2k955123







            • 1




              There's a typo in the last term (there should be a minus sign instead of a plus).
              – yoann
              Aug 5 at 10:10










            • @yoann Thanks for pointing out!!
              – Robert Z
              Aug 5 at 10:16












            • 1




              There's a typo in the last term (there should be a minus sign instead of a plus).
              – yoann
              Aug 5 at 10:10










            • @yoann Thanks for pointing out!!
              – Robert Z
              Aug 5 at 10:16







            1




            1




            There's a typo in the last term (there should be a minus sign instead of a plus).
            – yoann
            Aug 5 at 10:10




            There's a typo in the last term (there should be a minus sign instead of a plus).
            – yoann
            Aug 5 at 10:10












            @yoann Thanks for pointing out!!
            – Robert Z
            Aug 5 at 10:16




            @yoann Thanks for pointing out!!
            – Robert Z
            Aug 5 at 10:16










            up vote
            7
            down vote













            If you multiply your series by $r-1$, you get$$(r-1)+(r^2-1)s+(r^3-1)s^2+cdots,tag1$$which is the sum of$$r+r^2s+r^3s^2+cdots$$with$$-1-s-s^2-cdots$$The sum of the first series is $frac r1-rs$, whereas the sum of the second one is $-frac11-s$. Therefore,$$(1)=frac1r-1left(frac r1-rs-frac11-sright)=frac1(1-s)(1-rs).$$






            share|cite|improve this answer



















            • 1




              @farruhota I've edited my answer. Thank you.
              – José Carlos Santos
              Aug 5 at 9:36














            up vote
            7
            down vote













            If you multiply your series by $r-1$, you get$$(r-1)+(r^2-1)s+(r^3-1)s^2+cdots,tag1$$which is the sum of$$r+r^2s+r^3s^2+cdots$$with$$-1-s-s^2-cdots$$The sum of the first series is $frac r1-rs$, whereas the sum of the second one is $-frac11-s$. Therefore,$$(1)=frac1r-1left(frac r1-rs-frac11-sright)=frac1(1-s)(1-rs).$$






            share|cite|improve this answer



















            • 1




              @farruhota I've edited my answer. Thank you.
              – José Carlos Santos
              Aug 5 at 9:36












            up vote
            7
            down vote










            up vote
            7
            down vote









            If you multiply your series by $r-1$, you get$$(r-1)+(r^2-1)s+(r^3-1)s^2+cdots,tag1$$which is the sum of$$r+r^2s+r^3s^2+cdots$$with$$-1-s-s^2-cdots$$The sum of the first series is $frac r1-rs$, whereas the sum of the second one is $-frac11-s$. Therefore,$$(1)=frac1r-1left(frac r1-rs-frac11-sright)=frac1(1-s)(1-rs).$$






            share|cite|improve this answer















            If you multiply your series by $r-1$, you get$$(r-1)+(r^2-1)s+(r^3-1)s^2+cdots,tag1$$which is the sum of$$r+r^2s+r^3s^2+cdots$$with$$-1-s-s^2-cdots$$The sum of the first series is $frac r1-rs$, whereas the sum of the second one is $-frac11-s$. Therefore,$$(1)=frac1r-1left(frac r1-rs-frac11-sright)=frac1(1-s)(1-rs).$$







            share|cite|improve this answer















            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 5 at 9:36


























            answered Aug 5 at 8:20









            José Carlos Santos

            115k1698177




            115k1698177







            • 1




              @farruhota I've edited my answer. Thank you.
              – José Carlos Santos
              Aug 5 at 9:36












            • 1




              @farruhota I've edited my answer. Thank you.
              – José Carlos Santos
              Aug 5 at 9:36







            1




            1




            @farruhota I've edited my answer. Thank you.
            – José Carlos Santos
            Aug 5 at 9:36




            @farruhota I've edited my answer. Thank you.
            – José Carlos Santos
            Aug 5 at 9:36












             

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