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
    18 hours ago














up vote
7
down vote

favorite
4












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
    18 hours ago












up vote
7
down vote

favorite
4









up vote
7
down vote

favorite
4






4





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 3 hours ago









Asaf Karagila

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asked 18 hours ago









Loo Soo Yong

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624







  • 2




    Hint: $1+r+r^2 + cdots + r^k = frac1-r^k+11-r$
    – achille hui
    18 hours ago












  • 2




    Hint: $1+r+r^2 + cdots + r^k = frac1-r^k+11-r$
    – achille hui
    18 hours ago







2




2




Hint: $1+r+r^2 + cdots + r^k = frac1-r^k+11-r$
– achille hui
18 hours ago




Hint: $1+r+r^2 + cdots + r^k = frac1-r^k+11-r$
– achille hui
18 hours ago










3 Answers
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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
    17 hours ago










  • @yoann Thanks for pointing out!!
    – Robert Z
    16 hours ago

















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
    17 hours ago

















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3
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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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    3 Answers
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    3 Answers
    3






    active

    oldest

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    active

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    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
      17 hours ago










    • @yoann Thanks for pointing out!!
      – Robert Z
      16 hours ago














    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
      17 hours ago










    • @yoann Thanks for pointing out!!
      – Robert Z
      16 hours ago












    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 16 hours ago


























    answered 18 hours ago









    Robert Z

    83.2k953122




    83.2k953122







    • 1




      There's a typo in the last term (there should be a minus sign instead of a plus).
      – yoann
      17 hours ago










    • @yoann Thanks for pointing out!!
      – Robert Z
      16 hours ago












    • 1




      There's a typo in the last term (there should be a minus sign instead of a plus).
      – yoann
      17 hours ago










    • @yoann Thanks for pointing out!!
      – Robert Z
      16 hours ago







    1




    1




    There's a typo in the last term (there should be a minus sign instead of a plus).
    – yoann
    17 hours ago




    There's a typo in the last term (there should be a minus sign instead of a plus).
    – yoann
    17 hours ago












    @yoann Thanks for pointing out!!
    – Robert Z
    16 hours ago




    @yoann Thanks for pointing out!!
    – Robert Z
    16 hours ago










    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
      17 hours ago














    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
      17 hours ago












    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 17 hours ago


























    answered 18 hours ago









    José Carlos Santos

    111k1695171




    111k1695171







    • 1




      @farruhota I've edited my answer. Thank you.
      – José Carlos Santos
      17 hours ago












    • 1




      @farruhota I've edited my answer. Thank you.
      – José Carlos Santos
      17 hours ago







    1




    1




    @farruhota I've edited my answer. Thank you.
    – José Carlos Santos
    17 hours ago




    @farruhota I've edited my answer. Thank you.
    – José Carlos Santos
    17 hours ago










    up vote
    3
    down vote













    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
      3
      down vote













      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
        3
        down vote










        up vote
        3
        down vote









        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 17 hours ago









        farruhota

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