How is the partial sum of a geometric sequence calculated? [duplicate]

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3












$begingroup$



This question already has an answer here:



  • Formula for calculating $sum_n=0^mnr^n$

    4 answers



I have been asked to calculate a general formula for $1 + b + 2b^2 + 3b^3 + ... + Nb^N$



I think that a part of this is a geometric sequence, and I have rewritten this as



$f(x) = 1 + sum_i=1^n icdot x^i$



(I couldn't figure out a way to make the first term be 1 with the progression so I just removed it from the series instead)



When I look on Wolfram Alpha it says that the partial sum formula for $ sum_i=1^n icdot x^i$ is:



$$sum_i=1^n icdot x^i = frac(nx-n-1)x^n+1+x(1-x)^2$$



On this question, an answer said that the general formula for the sum of a finite geometric series is:



$$sum_k=0^n-1x^k = frac1-x^n1-x$$



But if I substitute my ($icdot x^i$) into the formula mentioned above I don't get the same value as the one Wolfram gives me.



Two questions:



  1. Is it correct to pull the first term out of the series so it becomes a geometric series, or is there another way?

  2. How did Wolfram Alpha calculate that expression?









share|cite|improve this question











$endgroup$



marked as duplicate by Hans Lundmark, Cesareo, Asaf Karagila Mar 18 at 8:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • 5




    $begingroup$
    The ratio of successive terms is not constant, so this is not a geometric sequence, though it may be related to one
    $endgroup$
    – J. W. Tanner
    Mar 17 at 13:54











  • $begingroup$
    Thank you @J.W.Tanner, I will edit the question
    $endgroup$
    – Joon
    Mar 17 at 13:59






  • 1




    $begingroup$
    Ok; by the way, I think we meant series (sum) rather than sequence
    $endgroup$
    – J. W. Tanner
    Mar 17 at 14:02






  • 2




    $begingroup$
    This page could be helpful: math.stackexchange.com/questions/2782812/finite-sum-kxk. Or you can let $S = x + 2x^2 + 3x^3 + cdots +nx^n$, and consider $S - xS$, and make sure you know the geometric series formula.
    $endgroup$
    – Minus One-Twelfth
    Mar 17 at 14:06
















3












$begingroup$



This question already has an answer here:



  • Formula for calculating $sum_n=0^mnr^n$

    4 answers



I have been asked to calculate a general formula for $1 + b + 2b^2 + 3b^3 + ... + Nb^N$



I think that a part of this is a geometric sequence, and I have rewritten this as



$f(x) = 1 + sum_i=1^n icdot x^i$



(I couldn't figure out a way to make the first term be 1 with the progression so I just removed it from the series instead)



When I look on Wolfram Alpha it says that the partial sum formula for $ sum_i=1^n icdot x^i$ is:



$$sum_i=1^n icdot x^i = frac(nx-n-1)x^n+1+x(1-x)^2$$



On this question, an answer said that the general formula for the sum of a finite geometric series is:



$$sum_k=0^n-1x^k = frac1-x^n1-x$$



But if I substitute my ($icdot x^i$) into the formula mentioned above I don't get the same value as the one Wolfram gives me.



Two questions:



  1. Is it correct to pull the first term out of the series so it becomes a geometric series, or is there another way?

  2. How did Wolfram Alpha calculate that expression?









share|cite|improve this question











$endgroup$



marked as duplicate by Hans Lundmark, Cesareo, Asaf Karagila Mar 18 at 8:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • 5




    $begingroup$
    The ratio of successive terms is not constant, so this is not a geometric sequence, though it may be related to one
    $endgroup$
    – J. W. Tanner
    Mar 17 at 13:54











  • $begingroup$
    Thank you @J.W.Tanner, I will edit the question
    $endgroup$
    – Joon
    Mar 17 at 13:59






  • 1




    $begingroup$
    Ok; by the way, I think we meant series (sum) rather than sequence
    $endgroup$
    – J. W. Tanner
    Mar 17 at 14:02






  • 2




    $begingroup$
    This page could be helpful: math.stackexchange.com/questions/2782812/finite-sum-kxk. Or you can let $S = x + 2x^2 + 3x^3 + cdots +nx^n$, and consider $S - xS$, and make sure you know the geometric series formula.
    $endgroup$
    – Minus One-Twelfth
    Mar 17 at 14:06














3












3








3


4



$begingroup$



This question already has an answer here:



  • Formula for calculating $sum_n=0^mnr^n$

    4 answers



I have been asked to calculate a general formula for $1 + b + 2b^2 + 3b^3 + ... + Nb^N$



I think that a part of this is a geometric sequence, and I have rewritten this as



$f(x) = 1 + sum_i=1^n icdot x^i$



(I couldn't figure out a way to make the first term be 1 with the progression so I just removed it from the series instead)



When I look on Wolfram Alpha it says that the partial sum formula for $ sum_i=1^n icdot x^i$ is:



$$sum_i=1^n icdot x^i = frac(nx-n-1)x^n+1+x(1-x)^2$$



On this question, an answer said that the general formula for the sum of a finite geometric series is:



$$sum_k=0^n-1x^k = frac1-x^n1-x$$



But if I substitute my ($icdot x^i$) into the formula mentioned above I don't get the same value as the one Wolfram gives me.



Two questions:



  1. Is it correct to pull the first term out of the series so it becomes a geometric series, or is there another way?

  2. How did Wolfram Alpha calculate that expression?









share|cite|improve this question











$endgroup$





This question already has an answer here:



  • Formula for calculating $sum_n=0^mnr^n$

    4 answers



I have been asked to calculate a general formula for $1 + b + 2b^2 + 3b^3 + ... + Nb^N$



I think that a part of this is a geometric sequence, and I have rewritten this as



$f(x) = 1 + sum_i=1^n icdot x^i$



(I couldn't figure out a way to make the first term be 1 with the progression so I just removed it from the series instead)



When I look on Wolfram Alpha it says that the partial sum formula for $ sum_i=1^n icdot x^i$ is:



$$sum_i=1^n icdot x^i = frac(nx-n-1)x^n+1+x(1-x)^2$$



On this question, an answer said that the general formula for the sum of a finite geometric series is:



$$sum_k=0^n-1x^k = frac1-x^n1-x$$



But if I substitute my ($icdot x^i$) into the formula mentioned above I don't get the same value as the one Wolfram gives me.



Two questions:



  1. Is it correct to pull the first term out of the series so it becomes a geometric series, or is there another way?

  2. How did Wolfram Alpha calculate that expression?




This question already has an answer here:



  • Formula for calculating $sum_n=0^mnr^n$

    4 answers







calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 17 at 14:51









Max

1,0041319




1,0041319










asked Mar 17 at 13:50









JoonJoon

1185




1185




marked as duplicate by Hans Lundmark, Cesareo, Asaf Karagila Mar 18 at 8:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Hans Lundmark, Cesareo, Asaf Karagila Mar 18 at 8:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









  • 5




    $begingroup$
    The ratio of successive terms is not constant, so this is not a geometric sequence, though it may be related to one
    $endgroup$
    – J. W. Tanner
    Mar 17 at 13:54











  • $begingroup$
    Thank you @J.W.Tanner, I will edit the question
    $endgroup$
    – Joon
    Mar 17 at 13:59






  • 1




    $begingroup$
    Ok; by the way, I think we meant series (sum) rather than sequence
    $endgroup$
    – J. W. Tanner
    Mar 17 at 14:02






  • 2




    $begingroup$
    This page could be helpful: math.stackexchange.com/questions/2782812/finite-sum-kxk. Or you can let $S = x + 2x^2 + 3x^3 + cdots +nx^n$, and consider $S - xS$, and make sure you know the geometric series formula.
    $endgroup$
    – Minus One-Twelfth
    Mar 17 at 14:06













  • 5




    $begingroup$
    The ratio of successive terms is not constant, so this is not a geometric sequence, though it may be related to one
    $endgroup$
    – J. W. Tanner
    Mar 17 at 13:54











  • $begingroup$
    Thank you @J.W.Tanner, I will edit the question
    $endgroup$
    – Joon
    Mar 17 at 13:59






  • 1




    $begingroup$
    Ok; by the way, I think we meant series (sum) rather than sequence
    $endgroup$
    – J. W. Tanner
    Mar 17 at 14:02






  • 2




    $begingroup$
    This page could be helpful: math.stackexchange.com/questions/2782812/finite-sum-kxk. Or you can let $S = x + 2x^2 + 3x^3 + cdots +nx^n$, and consider $S - xS$, and make sure you know the geometric series formula.
    $endgroup$
    – Minus One-Twelfth
    Mar 17 at 14:06








5




5




$begingroup$
The ratio of successive terms is not constant, so this is not a geometric sequence, though it may be related to one
$endgroup$
– J. W. Tanner
Mar 17 at 13:54





$begingroup$
The ratio of successive terms is not constant, so this is not a geometric sequence, though it may be related to one
$endgroup$
– J. W. Tanner
Mar 17 at 13:54













$begingroup$
Thank you @J.W.Tanner, I will edit the question
$endgroup$
– Joon
Mar 17 at 13:59




$begingroup$
Thank you @J.W.Tanner, I will edit the question
$endgroup$
– Joon
Mar 17 at 13:59




1




1




$begingroup$
Ok; by the way, I think we meant series (sum) rather than sequence
$endgroup$
– J. W. Tanner
Mar 17 at 14:02




$begingroup$
Ok; by the way, I think we meant series (sum) rather than sequence
$endgroup$
– J. W. Tanner
Mar 17 at 14:02




2




2




$begingroup$
This page could be helpful: math.stackexchange.com/questions/2782812/finite-sum-kxk. Or you can let $S = x + 2x^2 + 3x^3 + cdots +nx^n$, and consider $S - xS$, and make sure you know the geometric series formula.
$endgroup$
– Minus One-Twelfth
Mar 17 at 14:06





$begingroup$
This page could be helpful: math.stackexchange.com/questions/2782812/finite-sum-kxk. Or you can let $S = x + 2x^2 + 3x^3 + cdots +nx^n$, and consider $S - xS$, and make sure you know the geometric series formula.
$endgroup$
– Minus One-Twelfth
Mar 17 at 14:06











5 Answers
5






active

oldest

votes


















0












$begingroup$

Here's what I did to get the formula for partial sums of this series:





It was too much to type in LaTeX, so just did it on paper. Hope you don't mind.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thank you! Could you please explain how you know to add the (1 - x) / (1 - x) to the series?
    $endgroup$
    – Joon
    Mar 17 at 16:31






  • 1




    $begingroup$
    @Joon it's just a clever trick. There's really no reason why would one multiply and divide by (1-x). I guess it just works :)
    $endgroup$
    – KKZiomek
    Mar 17 at 17:15






  • 1




    $begingroup$
    If you analyze it closely you see that it cleverly cancels out like terms in a fashion that creates a geometric sequence. So I guess to know to multiply by (1-x) you have to observe and be clever. I didn't come up with this method. I found it in some paper about sequences long time ago
    $endgroup$
    – KKZiomek
    Mar 17 at 17:17






  • 1




    $begingroup$
    Thank you again, I could understand your method and follow along. Really appreciate it
    $endgroup$
    – Joon
    Mar 17 at 17:23



















11












$begingroup$

The extra factor of $i$ or $k$ or whatever you call it means we no longer have a geometric series, but we can get one by differentiating with respect to $x$. You'll want to check powers of $x$ carefully too.






share|cite|improve this answer











$endgroup$




















    6












    $begingroup$

    Your presumptions are wrong. The first series is not a geometric series. and substituting $i*x^i$ into $sum_k=0^N x^k$ is not the same as $sum_i=0^Ni*x^i$.



    Try to solve the problem using the principle of induction:



    Let us assume the above relation holds for $n$: $sum_i=0^n ix^i=frac(nx-n-1)x^n+1+x(1-x)^2equiv=A_n$.



    Now let us show it must also holds for $n+1$ and therefore for all $ninmathcalN$.
    Thus,



    beginalign
    sum_i=0^n+1ix^i&=underbracesum_i=0^nix^i_=A_n+(n+1)x^n+1\
    &=A_n+(n+1)x^n+1\
    &=frac(nx-n-1)x^n+1+x(1-x)^2+(n+1)x^n+1\
    &=frac(nx-n-1)x^n+1+x(1-x)^2+(n+1)x^n+1frac(1-x)^2(1-x)^2\
    &=frac(nx-n-1)x^n+1+x+(n+1)x^n+1*(x-1)^2(1-x)^2\
    &=frac1(1-x)^2biggl((nx-n-1)x^n+1+x+(n+1)x^n+1(1+x^2-2x)biggr)\
    &=frac1(1-x)^2biggl(x^n+2((n+1)x-n-2)+xbiggr)\
    &=frac1(1-x)^2biggl(x^(n+1)+1((n+1)x-(n+1)-1)+xbiggr)
    endalign



    So if it holds for any $n$ it must hold for $n+1$ too and therefore for all natural numbers. Now, you only have to check that it holds for $n=0$ or $n=1$ and you are done.






    share|cite|improve this answer











    $endgroup$




















      6












      $begingroup$

      Hint: Think derivatives w.r.t. $b,$:



      $$1 + b + 2b^2 + 3b^3 + dotsm + Nb^N=bsum_k=1^Nk,b^k-1=bBigl(sum_k=0^Nb^kBigr)^!!'$$






      share|cite|improve this answer











      $endgroup$




















        3












        $begingroup$

        Start with $$S=1 + b + 2b^2 + 3b^3 + cdots + Nb^N$$



        Multiply by $b$ and then subtract to give
        $$bS=b + b^2 + 2b^3 + 3b^4 + cdots +(N-1)b^N + Nb^N+1$$
        $$(b-1)S = -1 - b^2 - b^3 - b^4 - cdots -b^N + Nb^N+1$$



        Do that again
        $$b(b-1)S = -b - b^3 - b^4 - b^5 - cdots -b^N-b^N+1 - Nb^N+2$$
        $$(b-1)^2S = 1-b +b^2 -(N+1)b^N+1 + Nb^N+2$$



        Finally simplify
        $$S = dfrac1-b +b^2 -(N+1)b^N+1 + Nb^N+2(b-1)^2.$$



        As an example, with $b=2$ and $N=3$ the first line gives $S=1+2+8+24=35$ while the final line gives $S=frac1-2+4-64+961^2= 35$






        share|cite|improve this answer









        $endgroup$












        • $begingroup$
          Thank you for the understandable explanation!
          $endgroup$
          – Joon
          Mar 17 at 18:32

















        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        Here's what I did to get the formula for partial sums of this series:





        It was too much to type in LaTeX, so just did it on paper. Hope you don't mind.






        share|cite|improve this answer











        $endgroup$












        • $begingroup$
          Thank you! Could you please explain how you know to add the (1 - x) / (1 - x) to the series?
          $endgroup$
          – Joon
          Mar 17 at 16:31






        • 1




          $begingroup$
          @Joon it's just a clever trick. There's really no reason why would one multiply and divide by (1-x). I guess it just works :)
          $endgroup$
          – KKZiomek
          Mar 17 at 17:15






        • 1




          $begingroup$
          If you analyze it closely you see that it cleverly cancels out like terms in a fashion that creates a geometric sequence. So I guess to know to multiply by (1-x) you have to observe and be clever. I didn't come up with this method. I found it in some paper about sequences long time ago
          $endgroup$
          – KKZiomek
          Mar 17 at 17:17






        • 1




          $begingroup$
          Thank you again, I could understand your method and follow along. Really appreciate it
          $endgroup$
          – Joon
          Mar 17 at 17:23
















        0












        $begingroup$

        Here's what I did to get the formula for partial sums of this series:





        It was too much to type in LaTeX, so just did it on paper. Hope you don't mind.






        share|cite|improve this answer











        $endgroup$












        • $begingroup$
          Thank you! Could you please explain how you know to add the (1 - x) / (1 - x) to the series?
          $endgroup$
          – Joon
          Mar 17 at 16:31






        • 1




          $begingroup$
          @Joon it's just a clever trick. There's really no reason why would one multiply and divide by (1-x). I guess it just works :)
          $endgroup$
          – KKZiomek
          Mar 17 at 17:15






        • 1




          $begingroup$
          If you analyze it closely you see that it cleverly cancels out like terms in a fashion that creates a geometric sequence. So I guess to know to multiply by (1-x) you have to observe and be clever. I didn't come up with this method. I found it in some paper about sequences long time ago
          $endgroup$
          – KKZiomek
          Mar 17 at 17:17






        • 1




          $begingroup$
          Thank you again, I could understand your method and follow along. Really appreciate it
          $endgroup$
          – Joon
          Mar 17 at 17:23














        0












        0








        0





        $begingroup$

        Here's what I did to get the formula for partial sums of this series:





        It was too much to type in LaTeX, so just did it on paper. Hope you don't mind.






        share|cite|improve this answer











        $endgroup$



        Here's what I did to get the formula for partial sums of this series:





        It was too much to type in LaTeX, so just did it on paper. Hope you don't mind.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 17 at 14:23

























        answered Mar 17 at 14:02









        KKZiomekKKZiomek

        2,2441641




        2,2441641











        • $begingroup$
          Thank you! Could you please explain how you know to add the (1 - x) / (1 - x) to the series?
          $endgroup$
          – Joon
          Mar 17 at 16:31






        • 1




          $begingroup$
          @Joon it's just a clever trick. There's really no reason why would one multiply and divide by (1-x). I guess it just works :)
          $endgroup$
          – KKZiomek
          Mar 17 at 17:15






        • 1




          $begingroup$
          If you analyze it closely you see that it cleverly cancels out like terms in a fashion that creates a geometric sequence. So I guess to know to multiply by (1-x) you have to observe and be clever. I didn't come up with this method. I found it in some paper about sequences long time ago
          $endgroup$
          – KKZiomek
          Mar 17 at 17:17






        • 1




          $begingroup$
          Thank you again, I could understand your method and follow along. Really appreciate it
          $endgroup$
          – Joon
          Mar 17 at 17:23

















        • $begingroup$
          Thank you! Could you please explain how you know to add the (1 - x) / (1 - x) to the series?
          $endgroup$
          – Joon
          Mar 17 at 16:31






        • 1




          $begingroup$
          @Joon it's just a clever trick. There's really no reason why would one multiply and divide by (1-x). I guess it just works :)
          $endgroup$
          – KKZiomek
          Mar 17 at 17:15






        • 1




          $begingroup$
          If you analyze it closely you see that it cleverly cancels out like terms in a fashion that creates a geometric sequence. So I guess to know to multiply by (1-x) you have to observe and be clever. I didn't come up with this method. I found it in some paper about sequences long time ago
          $endgroup$
          – KKZiomek
          Mar 17 at 17:17






        • 1




          $begingroup$
          Thank you again, I could understand your method and follow along. Really appreciate it
          $endgroup$
          – Joon
          Mar 17 at 17:23
















        $begingroup$
        Thank you! Could you please explain how you know to add the (1 - x) / (1 - x) to the series?
        $endgroup$
        – Joon
        Mar 17 at 16:31




        $begingroup$
        Thank you! Could you please explain how you know to add the (1 - x) / (1 - x) to the series?
        $endgroup$
        – Joon
        Mar 17 at 16:31




        1




        1




        $begingroup$
        @Joon it's just a clever trick. There's really no reason why would one multiply and divide by (1-x). I guess it just works :)
        $endgroup$
        – KKZiomek
        Mar 17 at 17:15




        $begingroup$
        @Joon it's just a clever trick. There's really no reason why would one multiply and divide by (1-x). I guess it just works :)
        $endgroup$
        – KKZiomek
        Mar 17 at 17:15




        1




        1




        $begingroup$
        If you analyze it closely you see that it cleverly cancels out like terms in a fashion that creates a geometric sequence. So I guess to know to multiply by (1-x) you have to observe and be clever. I didn't come up with this method. I found it in some paper about sequences long time ago
        $endgroup$
        – KKZiomek
        Mar 17 at 17:17




        $begingroup$
        If you analyze it closely you see that it cleverly cancels out like terms in a fashion that creates a geometric sequence. So I guess to know to multiply by (1-x) you have to observe and be clever. I didn't come up with this method. I found it in some paper about sequences long time ago
        $endgroup$
        – KKZiomek
        Mar 17 at 17:17




        1




        1




        $begingroup$
        Thank you again, I could understand your method and follow along. Really appreciate it
        $endgroup$
        – Joon
        Mar 17 at 17:23





        $begingroup$
        Thank you again, I could understand your method and follow along. Really appreciate it
        $endgroup$
        – Joon
        Mar 17 at 17:23












        11












        $begingroup$

        The extra factor of $i$ or $k$ or whatever you call it means we no longer have a geometric series, but we can get one by differentiating with respect to $x$. You'll want to check powers of $x$ carefully too.






        share|cite|improve this answer











        $endgroup$

















          11












          $begingroup$

          The extra factor of $i$ or $k$ or whatever you call it means we no longer have a geometric series, but we can get one by differentiating with respect to $x$. You'll want to check powers of $x$ carefully too.






          share|cite|improve this answer











          $endgroup$















            11












            11








            11





            $begingroup$

            The extra factor of $i$ or $k$ or whatever you call it means we no longer have a geometric series, but we can get one by differentiating with respect to $x$. You'll want to check powers of $x$ carefully too.






            share|cite|improve this answer











            $endgroup$



            The extra factor of $i$ or $k$ or whatever you call it means we no longer have a geometric series, but we can get one by differentiating with respect to $x$. You'll want to check powers of $x$ carefully too.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 17 at 20:27

























            answered Mar 17 at 13:55









            J.G.J.G.

            33.5k23252




            33.5k23252





















                6












                $begingroup$

                Your presumptions are wrong. The first series is not a geometric series. and substituting $i*x^i$ into $sum_k=0^N x^k$ is not the same as $sum_i=0^Ni*x^i$.



                Try to solve the problem using the principle of induction:



                Let us assume the above relation holds for $n$: $sum_i=0^n ix^i=frac(nx-n-1)x^n+1+x(1-x)^2equiv=A_n$.



                Now let us show it must also holds for $n+1$ and therefore for all $ninmathcalN$.
                Thus,



                beginalign
                sum_i=0^n+1ix^i&=underbracesum_i=0^nix^i_=A_n+(n+1)x^n+1\
                &=A_n+(n+1)x^n+1\
                &=frac(nx-n-1)x^n+1+x(1-x)^2+(n+1)x^n+1\
                &=frac(nx-n-1)x^n+1+x(1-x)^2+(n+1)x^n+1frac(1-x)^2(1-x)^2\
                &=frac(nx-n-1)x^n+1+x+(n+1)x^n+1*(x-1)^2(1-x)^2\
                &=frac1(1-x)^2biggl((nx-n-1)x^n+1+x+(n+1)x^n+1(1+x^2-2x)biggr)\
                &=frac1(1-x)^2biggl(x^n+2((n+1)x-n-2)+xbiggr)\
                &=frac1(1-x)^2biggl(x^(n+1)+1((n+1)x-(n+1)-1)+xbiggr)
                endalign



                So if it holds for any $n$ it must hold for $n+1$ too and therefore for all natural numbers. Now, you only have to check that it holds for $n=0$ or $n=1$ and you are done.






                share|cite|improve this answer











                $endgroup$

















                  6












                  $begingroup$

                  Your presumptions are wrong. The first series is not a geometric series. and substituting $i*x^i$ into $sum_k=0^N x^k$ is not the same as $sum_i=0^Ni*x^i$.



                  Try to solve the problem using the principle of induction:



                  Let us assume the above relation holds for $n$: $sum_i=0^n ix^i=frac(nx-n-1)x^n+1+x(1-x)^2equiv=A_n$.



                  Now let us show it must also holds for $n+1$ and therefore for all $ninmathcalN$.
                  Thus,



                  beginalign
                  sum_i=0^n+1ix^i&=underbracesum_i=0^nix^i_=A_n+(n+1)x^n+1\
                  &=A_n+(n+1)x^n+1\
                  &=frac(nx-n-1)x^n+1+x(1-x)^2+(n+1)x^n+1\
                  &=frac(nx-n-1)x^n+1+x(1-x)^2+(n+1)x^n+1frac(1-x)^2(1-x)^2\
                  &=frac(nx-n-1)x^n+1+x+(n+1)x^n+1*(x-1)^2(1-x)^2\
                  &=frac1(1-x)^2biggl((nx-n-1)x^n+1+x+(n+1)x^n+1(1+x^2-2x)biggr)\
                  &=frac1(1-x)^2biggl(x^n+2((n+1)x-n-2)+xbiggr)\
                  &=frac1(1-x)^2biggl(x^(n+1)+1((n+1)x-(n+1)-1)+xbiggr)
                  endalign



                  So if it holds for any $n$ it must hold for $n+1$ too and therefore for all natural numbers. Now, you only have to check that it holds for $n=0$ or $n=1$ and you are done.






                  share|cite|improve this answer











                  $endgroup$















                    6












                    6








                    6





                    $begingroup$

                    Your presumptions are wrong. The first series is not a geometric series. and substituting $i*x^i$ into $sum_k=0^N x^k$ is not the same as $sum_i=0^Ni*x^i$.



                    Try to solve the problem using the principle of induction:



                    Let us assume the above relation holds for $n$: $sum_i=0^n ix^i=frac(nx-n-1)x^n+1+x(1-x)^2equiv=A_n$.



                    Now let us show it must also holds for $n+1$ and therefore for all $ninmathcalN$.
                    Thus,



                    beginalign
                    sum_i=0^n+1ix^i&=underbracesum_i=0^nix^i_=A_n+(n+1)x^n+1\
                    &=A_n+(n+1)x^n+1\
                    &=frac(nx-n-1)x^n+1+x(1-x)^2+(n+1)x^n+1\
                    &=frac(nx-n-1)x^n+1+x(1-x)^2+(n+1)x^n+1frac(1-x)^2(1-x)^2\
                    &=frac(nx-n-1)x^n+1+x+(n+1)x^n+1*(x-1)^2(1-x)^2\
                    &=frac1(1-x)^2biggl((nx-n-1)x^n+1+x+(n+1)x^n+1(1+x^2-2x)biggr)\
                    &=frac1(1-x)^2biggl(x^n+2((n+1)x-n-2)+xbiggr)\
                    &=frac1(1-x)^2biggl(x^(n+1)+1((n+1)x-(n+1)-1)+xbiggr)
                    endalign



                    So if it holds for any $n$ it must hold for $n+1$ too and therefore for all natural numbers. Now, you only have to check that it holds for $n=0$ or $n=1$ and you are done.






                    share|cite|improve this answer











                    $endgroup$



                    Your presumptions are wrong. The first series is not a geometric series. and substituting $i*x^i$ into $sum_k=0^N x^k$ is not the same as $sum_i=0^Ni*x^i$.



                    Try to solve the problem using the principle of induction:



                    Let us assume the above relation holds for $n$: $sum_i=0^n ix^i=frac(nx-n-1)x^n+1+x(1-x)^2equiv=A_n$.



                    Now let us show it must also holds for $n+1$ and therefore for all $ninmathcalN$.
                    Thus,



                    beginalign
                    sum_i=0^n+1ix^i&=underbracesum_i=0^nix^i_=A_n+(n+1)x^n+1\
                    &=A_n+(n+1)x^n+1\
                    &=frac(nx-n-1)x^n+1+x(1-x)^2+(n+1)x^n+1\
                    &=frac(nx-n-1)x^n+1+x(1-x)^2+(n+1)x^n+1frac(1-x)^2(1-x)^2\
                    &=frac(nx-n-1)x^n+1+x+(n+1)x^n+1*(x-1)^2(1-x)^2\
                    &=frac1(1-x)^2biggl((nx-n-1)x^n+1+x+(n+1)x^n+1(1+x^2-2x)biggr)\
                    &=frac1(1-x)^2biggl(x^n+2((n+1)x-n-2)+xbiggr)\
                    &=frac1(1-x)^2biggl(x^(n+1)+1((n+1)x-(n+1)-1)+xbiggr)
                    endalign



                    So if it holds for any $n$ it must hold for $n+1$ too and therefore for all natural numbers. Now, you only have to check that it holds for $n=0$ or $n=1$ and you are done.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Mar 17 at 17:56

























                    answered Mar 17 at 14:27









                    Gkhan CebsGkhan Cebs

                    813




                    813





















                        6












                        $begingroup$

                        Hint: Think derivatives w.r.t. $b,$:



                        $$1 + b + 2b^2 + 3b^3 + dotsm + Nb^N=bsum_k=1^Nk,b^k-1=bBigl(sum_k=0^Nb^kBigr)^!!'$$






                        share|cite|improve this answer











                        $endgroup$

















                          6












                          $begingroup$

                          Hint: Think derivatives w.r.t. $b,$:



                          $$1 + b + 2b^2 + 3b^3 + dotsm + Nb^N=bsum_k=1^Nk,b^k-1=bBigl(sum_k=0^Nb^kBigr)^!!'$$






                          share|cite|improve this answer











                          $endgroup$















                            6












                            6








                            6





                            $begingroup$

                            Hint: Think derivatives w.r.t. $b,$:



                            $$1 + b + 2b^2 + 3b^3 + dotsm + Nb^N=bsum_k=1^Nk,b^k-1=bBigl(sum_k=0^Nb^kBigr)^!!'$$






                            share|cite|improve this answer











                            $endgroup$



                            Hint: Think derivatives w.r.t. $b,$:



                            $$1 + b + 2b^2 + 3b^3 + dotsm + Nb^N=bsum_k=1^Nk,b^k-1=bBigl(sum_k=0^Nb^kBigr)^!!'$$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Mar 18 at 0:44

























                            answered Mar 17 at 14:06









                            BernardBernard

                            124k742117




                            124k742117





















                                3












                                $begingroup$

                                Start with $$S=1 + b + 2b^2 + 3b^3 + cdots + Nb^N$$



                                Multiply by $b$ and then subtract to give
                                $$bS=b + b^2 + 2b^3 + 3b^4 + cdots +(N-1)b^N + Nb^N+1$$
                                $$(b-1)S = -1 - b^2 - b^3 - b^4 - cdots -b^N + Nb^N+1$$



                                Do that again
                                $$b(b-1)S = -b - b^3 - b^4 - b^5 - cdots -b^N-b^N+1 - Nb^N+2$$
                                $$(b-1)^2S = 1-b +b^2 -(N+1)b^N+1 + Nb^N+2$$



                                Finally simplify
                                $$S = dfrac1-b +b^2 -(N+1)b^N+1 + Nb^N+2(b-1)^2.$$



                                As an example, with $b=2$ and $N=3$ the first line gives $S=1+2+8+24=35$ while the final line gives $S=frac1-2+4-64+961^2= 35$






                                share|cite|improve this answer









                                $endgroup$












                                • $begingroup$
                                  Thank you for the understandable explanation!
                                  $endgroup$
                                  – Joon
                                  Mar 17 at 18:32















                                3












                                $begingroup$

                                Start with $$S=1 + b + 2b^2 + 3b^3 + cdots + Nb^N$$



                                Multiply by $b$ and then subtract to give
                                $$bS=b + b^2 + 2b^3 + 3b^4 + cdots +(N-1)b^N + Nb^N+1$$
                                $$(b-1)S = -1 - b^2 - b^3 - b^4 - cdots -b^N + Nb^N+1$$



                                Do that again
                                $$b(b-1)S = -b - b^3 - b^4 - b^5 - cdots -b^N-b^N+1 - Nb^N+2$$
                                $$(b-1)^2S = 1-b +b^2 -(N+1)b^N+1 + Nb^N+2$$



                                Finally simplify
                                $$S = dfrac1-b +b^2 -(N+1)b^N+1 + Nb^N+2(b-1)^2.$$



                                As an example, with $b=2$ and $N=3$ the first line gives $S=1+2+8+24=35$ while the final line gives $S=frac1-2+4-64+961^2= 35$






                                share|cite|improve this answer









                                $endgroup$












                                • $begingroup$
                                  Thank you for the understandable explanation!
                                  $endgroup$
                                  – Joon
                                  Mar 17 at 18:32













                                3












                                3








                                3





                                $begingroup$

                                Start with $$S=1 + b + 2b^2 + 3b^3 + cdots + Nb^N$$



                                Multiply by $b$ and then subtract to give
                                $$bS=b + b^2 + 2b^3 + 3b^4 + cdots +(N-1)b^N + Nb^N+1$$
                                $$(b-1)S = -1 - b^2 - b^3 - b^4 - cdots -b^N + Nb^N+1$$



                                Do that again
                                $$b(b-1)S = -b - b^3 - b^4 - b^5 - cdots -b^N-b^N+1 - Nb^N+2$$
                                $$(b-1)^2S = 1-b +b^2 -(N+1)b^N+1 + Nb^N+2$$



                                Finally simplify
                                $$S = dfrac1-b +b^2 -(N+1)b^N+1 + Nb^N+2(b-1)^2.$$



                                As an example, with $b=2$ and $N=3$ the first line gives $S=1+2+8+24=35$ while the final line gives $S=frac1-2+4-64+961^2= 35$






                                share|cite|improve this answer









                                $endgroup$



                                Start with $$S=1 + b + 2b^2 + 3b^3 + cdots + Nb^N$$



                                Multiply by $b$ and then subtract to give
                                $$bS=b + b^2 + 2b^3 + 3b^4 + cdots +(N-1)b^N + Nb^N+1$$
                                $$(b-1)S = -1 - b^2 - b^3 - b^4 - cdots -b^N + Nb^N+1$$



                                Do that again
                                $$b(b-1)S = -b - b^3 - b^4 - b^5 - cdots -b^N-b^N+1 - Nb^N+2$$
                                $$(b-1)^2S = 1-b +b^2 -(N+1)b^N+1 + Nb^N+2$$



                                Finally simplify
                                $$S = dfrac1-b +b^2 -(N+1)b^N+1 + Nb^N+2(b-1)^2.$$



                                As an example, with $b=2$ and $N=3$ the first line gives $S=1+2+8+24=35$ while the final line gives $S=frac1-2+4-64+961^2= 35$







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Mar 17 at 18:04









                                HenryHenry

                                101k482170




                                101k482170











                                • $begingroup$
                                  Thank you for the understandable explanation!
                                  $endgroup$
                                  – Joon
                                  Mar 17 at 18:32
















                                • $begingroup$
                                  Thank you for the understandable explanation!
                                  $endgroup$
                                  – Joon
                                  Mar 17 at 18:32















                                $begingroup$
                                Thank you for the understandable explanation!
                                $endgroup$
                                – Joon
                                Mar 17 at 18:32




                                $begingroup$
                                Thank you for the understandable explanation!
                                $endgroup$
                                – Joon
                                Mar 17 at 18:32


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