Computing $sum_{i=1}^k{icdot c^i}$ [duplicate]












1












$begingroup$



This question already has an answer here:




  • Formula for calculating $sum_{n=0}^{m}nr^n$

    4 answers





I would like to compute the following sum :
$$sum_{i=1}^k{icdot c^i}$$
where $k$ and $c$ are any real numbers.




I know how to compute $i$ and $c^i$ separately but I don't know how to do it when they are multiplied together.



Thanks a lot for the help!






summation







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marked as duplicate by Hans Lundmark, Community♦ Jan 27 at 21:06


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.


















  • $begingroup$
    Welcome to Maths SX! I suppose you mean you want to compute this sum?
    $endgroup$
    – Bernard
    Jan 27 at 18:23










  • $begingroup$
    Oh yes thank you !
    $endgroup$
    – Louis-Simon Cyr
    Jan 27 at 19:31
















1












$begingroup$



This question already has an answer here:




  • Formula for calculating $sum_{n=0}^{m}nr^n$

    4 answers





I would like to compute the following sum :
$$sum_{i=1}^k{icdot c^i}$$
where $k$ and $c$ are any real numbers.




I know how to compute $i$ and $c^i$ separately but I don't know how to do it when they are multiplied together.



Thanks a lot for the help!






summation







share|cite|improve this question











$endgroup$



marked as duplicate by Hans Lundmark, Community♦ Jan 27 at 21:06


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.


















  • $begingroup$
    Welcome to Maths SX! I suppose you mean you want to compute this sum?
    $endgroup$
    – Bernard
    Jan 27 at 18:23










  • $begingroup$
    Oh yes thank you !
    $endgroup$
    – Louis-Simon Cyr
    Jan 27 at 19:31














1












1








1





$begingroup$



This question already has an answer here:




  • Formula for calculating $sum_{n=0}^{m}nr^n$

    4 answers





I would like to compute the following sum :
$$sum_{i=1}^k{icdot c^i}$$
where $k$ and $c$ are any real numbers.




I know how to compute $i$ and $c^i$ separately but I don't know how to do it when they are multiplied together.



Thanks a lot for the help!






summation







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Formula for calculating $sum_{n=0}^{m}nr^n$

    4 answers





I would like to compute the following sum :
$$sum_{i=1}^k{icdot c^i}$$
where $k$ and $c$ are any real numbers.




I know how to compute $i$ and $c^i$ separately but I don't know how to do it when they are multiplied together.



Thanks a lot for the help!





This question already has an answer here:




  • Formula for calculating $sum_{n=0}^{m}nr^n$

    4 answers







summation




summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 27 at 20:11









Blue

49.2k870157




49.2k870157










asked Jan 27 at 18:20









Louis-Simon CyrLouis-Simon Cyr

85




85




marked as duplicate by Hans Lundmark, Community♦ Jan 27 at 21:06


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, Community♦ Jan 27 at 21:06


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.














  • $begingroup$
    Welcome to Maths SX! I suppose you mean you want to compute this sum?
    $endgroup$
    – Bernard
    Jan 27 at 18:23










  • $begingroup$
    Oh yes thank you !
    $endgroup$
    – Louis-Simon Cyr
    Jan 27 at 19:31


















  • $begingroup$
    Welcome to Maths SX! I suppose you mean you want to compute this sum?
    $endgroup$
    – Bernard
    Jan 27 at 18:23










  • $begingroup$
    Oh yes thank you !
    $endgroup$
    – Louis-Simon Cyr
    Jan 27 at 19:31
















$begingroup$
Welcome to Maths SX! I suppose you mean you want to compute this sum?
$endgroup$
– Bernard
Jan 27 at 18:23




$begingroup$
Welcome to Maths SX! I suppose you mean you want to compute this sum?
$endgroup$
– Bernard
Jan 27 at 18:23












$begingroup$
Oh yes thank you !
$endgroup$
– Louis-Simon Cyr
Jan 27 at 19:31




$begingroup$
Oh yes thank you !
$endgroup$
– Louis-Simon Cyr
Jan 27 at 19:31










3 Answers
3






active

oldest

votes


















2












$begingroup$

Hint:



Do some analysis:
$$sum_{i=1}^k ic^i=csum_{i=1}^kic^{i-1}=cBigl(sum_{i=0}^k c^iBigr)'. $$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Let $Z=sum_{i=1}^kc^i$, then (derivative with respect to $c$), $Z'=sum_{i=1}^kic^{i-1}$. When multiplied with $c$, we get your question, i.e. we want $cZ'$. You can easily substitute $Z=frac{1-c^{k+1}}{1-c}-1$.






    share|cite|improve this answer









    $endgroup$





















      -1












      $begingroup$

      Hint: Let $a,q$ be real numbers with $qne 1$:
      $$a + aq + aq^2 + ldots +aq^{n-1} = afrac{q^n-1}{q-1}.$$






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        This is a true fact, but it is irrelevant to the question. OP is trying to compute the sum $$1c^1 + 2 c^2 + 3 c^3 + cdots + k c^k$$ The coefficients on the powers of $c$ are not constant.
        $endgroup$
        – Blue
        Jan 27 at 20:15




















      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Hint:



      Do some analysis:
      $$sum_{i=1}^k ic^i=csum_{i=1}^kic^{i-1}=cBigl(sum_{i=0}^k c^iBigr)'. $$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Hint:



        Do some analysis:
        $$sum_{i=1}^k ic^i=csum_{i=1}^kic^{i-1}=cBigl(sum_{i=0}^k c^iBigr)'. $$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Hint:



          Do some analysis:
          $$sum_{i=1}^k ic^i=csum_{i=1}^kic^{i-1}=cBigl(sum_{i=0}^k c^iBigr)'. $$






          share|cite|improve this answer









          $endgroup$



          Hint:



          Do some analysis:
          $$sum_{i=1}^k ic^i=csum_{i=1}^kic^{i-1}=cBigl(sum_{i=0}^k c^iBigr)'. $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 27 at 18:27









          BernardBernard

          123k741117




          123k741117























              1












              $begingroup$

              Let $Z=sum_{i=1}^kc^i$, then (derivative with respect to $c$), $Z'=sum_{i=1}^kic^{i-1}$. When multiplied with $c$, we get your question, i.e. we want $cZ'$. You can easily substitute $Z=frac{1-c^{k+1}}{1-c}-1$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Let $Z=sum_{i=1}^kc^i$, then (derivative with respect to $c$), $Z'=sum_{i=1}^kic^{i-1}$. When multiplied with $c$, we get your question, i.e. we want $cZ'$. You can easily substitute $Z=frac{1-c^{k+1}}{1-c}-1$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let $Z=sum_{i=1}^kc^i$, then (derivative with respect to $c$), $Z'=sum_{i=1}^kic^{i-1}$. When multiplied with $c$, we get your question, i.e. we want $cZ'$. You can easily substitute $Z=frac{1-c^{k+1}}{1-c}-1$.






                  share|cite|improve this answer









                  $endgroup$



                  Let $Z=sum_{i=1}^kc^i$, then (derivative with respect to $c$), $Z'=sum_{i=1}^kic^{i-1}$. When multiplied with $c$, we get your question, i.e. we want $cZ'$. You can easily substitute $Z=frac{1-c^{k+1}}{1-c}-1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 27 at 18:28









                  gunesgunes

                  3747




                  3747























                      -1












                      $begingroup$

                      Hint: Let $a,q$ be real numbers with $qne 1$:
                      $$a + aq + aq^2 + ldots +aq^{n-1} = afrac{q^n-1}{q-1}.$$






                      share|cite|improve this answer









                      $endgroup$









                      • 1




                        $begingroup$
                        This is a true fact, but it is irrelevant to the question. OP is trying to compute the sum $$1c^1 + 2 c^2 + 3 c^3 + cdots + k c^k$$ The coefficients on the powers of $c$ are not constant.
                        $endgroup$
                        – Blue
                        Jan 27 at 20:15


















                      -1












                      $begingroup$

                      Hint: Let $a,q$ be real numbers with $qne 1$:
                      $$a + aq + aq^2 + ldots +aq^{n-1} = afrac{q^n-1}{q-1}.$$






                      share|cite|improve this answer









                      $endgroup$









                      • 1




                        $begingroup$
                        This is a true fact, but it is irrelevant to the question. OP is trying to compute the sum $$1c^1 + 2 c^2 + 3 c^3 + cdots + k c^k$$ The coefficients on the powers of $c$ are not constant.
                        $endgroup$
                        – Blue
                        Jan 27 at 20:15
















                      -1












                      -1








                      -1





                      $begingroup$

                      Hint: Let $a,q$ be real numbers with $qne 1$:
                      $$a + aq + aq^2 + ldots +aq^{n-1} = afrac{q^n-1}{q-1}.$$






                      share|cite|improve this answer









                      $endgroup$



                      Hint: Let $a,q$ be real numbers with $qne 1$:
                      $$a + aq + aq^2 + ldots +aq^{n-1} = afrac{q^n-1}{q-1}.$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 27 at 18:29









                      WuestenfuxWuestenfux

                      5,3331513




                      5,3331513








                      • 1




                        $begingroup$
                        This is a true fact, but it is irrelevant to the question. OP is trying to compute the sum $$1c^1 + 2 c^2 + 3 c^3 + cdots + k c^k$$ The coefficients on the powers of $c$ are not constant.
                        $endgroup$
                        – Blue
                        Jan 27 at 20:15
















                      • 1




                        $begingroup$
                        This is a true fact, but it is irrelevant to the question. OP is trying to compute the sum $$1c^1 + 2 c^2 + 3 c^3 + cdots + k c^k$$ The coefficients on the powers of $c$ are not constant.
                        $endgroup$
                        – Blue
                        Jan 27 at 20:15










                      1




                      1




                      $begingroup$
                      This is a true fact, but it is irrelevant to the question. OP is trying to compute the sum $$1c^1 + 2 c^2 + 3 c^3 + cdots + k c^k$$ The coefficients on the powers of $c$ are not constant.
                      $endgroup$
                      – Blue
                      Jan 27 at 20:15






                      $begingroup$
                      This is a true fact, but it is irrelevant to the question. OP is trying to compute the sum $$1c^1 + 2 c^2 + 3 c^3 + cdots + k c^k$$ The coefficients on the powers of $c$ are not constant.
                      $endgroup$
                      – Blue
                      Jan 27 at 20:15





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