Property of summation












0












$begingroup$


Very short question. Could you please explain me why



$$sum_{i=0}^{n-1} a = na$$
with $a$ a constant?
I know that



$$sum_{i=1}^{n} a = na$$



but in my case the sum starts from zero and finishes for $(n-1)$.



Thanks.






summation







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
    $endgroup$
    – Yves Daoust
    yesterday












  • $begingroup$
    Recognize that ${0,1,2,3,4,dots,n-1}$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements ${1,2,3,dots,n-1}$ and also the one additional zero element ${0}$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
    $endgroup$
    – JMoravitz
    yesterday










  • $begingroup$
    Thanks JMoravitz, your rationale was what I needed to convince myself.
    $endgroup$
    – Kolmogorovwannabe
    yesterday
















0












$begingroup$


Very short question. Could you please explain me why



$$sum_{i=0}^{n-1} a = na$$
with $a$ a constant?
I know that



$$sum_{i=1}^{n} a = na$$



but in my case the sum starts from zero and finishes for $(n-1)$.



Thanks.






summation







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
    $endgroup$
    – Yves Daoust
    yesterday












  • $begingroup$
    Recognize that ${0,1,2,3,4,dots,n-1}$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements ${1,2,3,dots,n-1}$ and also the one additional zero element ${0}$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
    $endgroup$
    – JMoravitz
    yesterday










  • $begingroup$
    Thanks JMoravitz, your rationale was what I needed to convince myself.
    $endgroup$
    – Kolmogorovwannabe
    yesterday














0












0








0





$begingroup$


Very short question. Could you please explain me why



$$sum_{i=0}^{n-1} a = na$$
with $a$ a constant?
I know that



$$sum_{i=1}^{n} a = na$$



but in my case the sum starts from zero and finishes for $(n-1)$.



Thanks.






summation







share|cite|improve this question











$endgroup$




Very short question. Could you please explain me why



$$sum_{i=0}^{n-1} a = na$$
with $a$ a constant?
I know that



$$sum_{i=1}^{n} a = na$$



but in my case the sum starts from zero and finishes for $(n-1)$.



Thanks.






summation




summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









MarianD

1,2791616




1,2791616










asked yesterday









KolmogorovwannabeKolmogorovwannabe

257




257








  • 7




    $begingroup$
    Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
    $endgroup$
    – Yves Daoust
    yesterday












  • $begingroup$
    Recognize that ${0,1,2,3,4,dots,n-1}$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements ${1,2,3,dots,n-1}$ and also the one additional zero element ${0}$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
    $endgroup$
    – JMoravitz
    yesterday










  • $begingroup$
    Thanks JMoravitz, your rationale was what I needed to convince myself.
    $endgroup$
    – Kolmogorovwannabe
    yesterday














  • 7




    $begingroup$
    Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
    $endgroup$
    – Yves Daoust
    yesterday












  • $begingroup$
    Recognize that ${0,1,2,3,4,dots,n-1}$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements ${1,2,3,dots,n-1}$ and also the one additional zero element ${0}$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
    $endgroup$
    – JMoravitz
    yesterday










  • $begingroup$
    Thanks JMoravitz, your rationale was what I needed to convince myself.
    $endgroup$
    – Kolmogorovwannabe
    yesterday








7




7




$begingroup$
Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
$endgroup$
– Yves Daoust
yesterday






$begingroup$
Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
$endgroup$
– Yves Daoust
yesterday














$begingroup$
Recognize that ${0,1,2,3,4,dots,n-1}$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements ${1,2,3,dots,n-1}$ and also the one additional zero element ${0}$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
$endgroup$
– JMoravitz
yesterday




$begingroup$
Recognize that ${0,1,2,3,4,dots,n-1}$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements ${1,2,3,dots,n-1}$ and also the one additional zero element ${0}$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
$endgroup$
– JMoravitz
yesterday












$begingroup$
Thanks JMoravitz, your rationale was what I needed to convince myself.
$endgroup$
– Kolmogorovwannabe
yesterday




$begingroup$
Thanks JMoravitz, your rationale was what I needed to convince myself.
$endgroup$
– Kolmogorovwannabe
yesterday










6 Answers
6






active

oldest

votes


















8












$begingroup$

Since you are allready convinced that $sum_{i=1}^{n}a=na$ this might help:



$sum_{i=0}^{n-1}a=a+sum_{i=1}^{n-1}a=a+sum_{i=1}^{n}a-a=sum_{i=1}^{n}a$






share|cite|improve this answer









$endgroup$





















    22












    $begingroup$

    In both cases - $$sum_{i=0}^{n-1}aquad text { and }quadsum_{i=1}^{n}a$$



    - there are exactly $n$ summands.



    enter image description here






    share|cite|improve this answer









    $endgroup$





















      3












      $begingroup$

      Since $a$ is not $i$-depending one can write: $$sum_{i=0}^{n-1}{a}=asum_{i=0}^{n-1}{1}$$
      And $sum_{i=0}^{n-1}{1}=1+1+cdots+1$ $n$ times which obviously is $n$.






      share|cite|improve this answer











      $endgroup$





















        2












        $begingroup$

        What is the definition of $sum_{i=0}^{n-1} x_i$? It is exactly $x_0+x_1+...x_{n-1}$. If $x_0=x_1=...x_{n-1}=a$ then it means you just sum $a$ $n$ times. And that gives $na$.






        share|cite|improve this answer











        $endgroup$





















          1












          $begingroup$

          You are summing $n$ terms, all equal to $a$. So
          $$sum_{i=0}^{n-1} a = a+ a+dotsb + a = na.$$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Hint:



            $a$ times the number of terms.






            share|cite|improve this answer









            $endgroup$













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              6 Answers
              6






              active

              oldest

              votes








              6 Answers
              6






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              8












              $begingroup$

              Since you are allready convinced that $sum_{i=1}^{n}a=na$ this might help:



              $sum_{i=0}^{n-1}a=a+sum_{i=1}^{n-1}a=a+sum_{i=1}^{n}a-a=sum_{i=1}^{n}a$






              share|cite|improve this answer









              $endgroup$


















                8












                $begingroup$

                Since you are allready convinced that $sum_{i=1}^{n}a=na$ this might help:



                $sum_{i=0}^{n-1}a=a+sum_{i=1}^{n-1}a=a+sum_{i=1}^{n}a-a=sum_{i=1}^{n}a$






                share|cite|improve this answer









                $endgroup$
















                  8












                  8








                  8





                  $begingroup$

                  Since you are allready convinced that $sum_{i=1}^{n}a=na$ this might help:



                  $sum_{i=0}^{n-1}a=a+sum_{i=1}^{n-1}a=a+sum_{i=1}^{n}a-a=sum_{i=1}^{n}a$






                  share|cite|improve this answer









                  $endgroup$



                  Since you are allready convinced that $sum_{i=1}^{n}a=na$ this might help:



                  $sum_{i=0}^{n-1}a=a+sum_{i=1}^{n-1}a=a+sum_{i=1}^{n}a-a=sum_{i=1}^{n}a$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  drhabdrhab

                  103k545136




                  103k545136























                      22












                      $begingroup$

                      In both cases - $$sum_{i=0}^{n-1}aquad text { and }quadsum_{i=1}^{n}a$$



                      - there are exactly $n$ summands.



                      enter image description here






                      share|cite|improve this answer









                      $endgroup$


















                        22












                        $begingroup$

                        In both cases - $$sum_{i=0}^{n-1}aquad text { and }quadsum_{i=1}^{n}a$$



                        - there are exactly $n$ summands.



                        enter image description here






                        share|cite|improve this answer









                        $endgroup$
















                          22












                          22








                          22





                          $begingroup$

                          In both cases - $$sum_{i=0}^{n-1}aquad text { and }quadsum_{i=1}^{n}a$$



                          - there are exactly $n$ summands.



                          enter image description here






                          share|cite|improve this answer









                          $endgroup$



                          In both cases - $$sum_{i=0}^{n-1}aquad text { and }quadsum_{i=1}^{n}a$$



                          - there are exactly $n$ summands.



                          enter image description here







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered yesterday









                          MarianDMarianD

                          1,2791616




                          1,2791616























                              3












                              $begingroup$

                              Since $a$ is not $i$-depending one can write: $$sum_{i=0}^{n-1}{a}=asum_{i=0}^{n-1}{1}$$
                              And $sum_{i=0}^{n-1}{1}=1+1+cdots+1$ $n$ times which obviously is $n$.






                              share|cite|improve this answer











                              $endgroup$


















                                3












                                $begingroup$

                                Since $a$ is not $i$-depending one can write: $$sum_{i=0}^{n-1}{a}=asum_{i=0}^{n-1}{1}$$
                                And $sum_{i=0}^{n-1}{1}=1+1+cdots+1$ $n$ times which obviously is $n$.






                                share|cite|improve this answer











                                $endgroup$
















                                  3












                                  3








                                  3





                                  $begingroup$

                                  Since $a$ is not $i$-depending one can write: $$sum_{i=0}^{n-1}{a}=asum_{i=0}^{n-1}{1}$$
                                  And $sum_{i=0}^{n-1}{1}=1+1+cdots+1$ $n$ times which obviously is $n$.






                                  share|cite|improve this answer











                                  $endgroup$



                                  Since $a$ is not $i$-depending one can write: $$sum_{i=0}^{n-1}{a}=asum_{i=0}^{n-1}{1}$$
                                  And $sum_{i=0}^{n-1}{1}=1+1+cdots+1$ $n$ times which obviously is $n$.







                                  share|cite|improve this answer














                                  share|cite|improve this answer



                                  share|cite|improve this answer








                                  edited yesterday









                                  MarianD

                                  1,2791616




                                  1,2791616










                                  answered yesterday









                                  HAMIDINE SOUMAREHAMIDINE SOUMARE

                                  1,05429




                                  1,05429























                                      2












                                      $begingroup$

                                      What is the definition of $sum_{i=0}^{n-1} x_i$? It is exactly $x_0+x_1+...x_{n-1}$. If $x_0=x_1=...x_{n-1}=a$ then it means you just sum $a$ $n$ times. And that gives $na$.






                                      share|cite|improve this answer











                                      $endgroup$


















                                        2












                                        $begingroup$

                                        What is the definition of $sum_{i=0}^{n-1} x_i$? It is exactly $x_0+x_1+...x_{n-1}$. If $x_0=x_1=...x_{n-1}=a$ then it means you just sum $a$ $n$ times. And that gives $na$.






                                        share|cite|improve this answer











                                        $endgroup$
















                                          2












                                          2








                                          2





                                          $begingroup$

                                          What is the definition of $sum_{i=0}^{n-1} x_i$? It is exactly $x_0+x_1+...x_{n-1}$. If $x_0=x_1=...x_{n-1}=a$ then it means you just sum $a$ $n$ times. And that gives $na$.






                                          share|cite|improve this answer











                                          $endgroup$



                                          What is the definition of $sum_{i=0}^{n-1} x_i$? It is exactly $x_0+x_1+...x_{n-1}$. If $x_0=x_1=...x_{n-1}=a$ then it means you just sum $a$ $n$ times. And that gives $na$.







                                          share|cite|improve this answer














                                          share|cite|improve this answer



                                          share|cite|improve this answer








                                          edited yesterday

























                                          answered yesterday









                                          MarkMark

                                          10.1k622




                                          10.1k622























                                              1












                                              $begingroup$

                                              You are summing $n$ terms, all equal to $a$. So
                                              $$sum_{i=0}^{n-1} a = a+ a+dotsb + a = na.$$






                                              share|cite|improve this answer









                                              $endgroup$


















                                                1












                                                $begingroup$

                                                You are summing $n$ terms, all equal to $a$. So
                                                $$sum_{i=0}^{n-1} a = a+ a+dotsb + a = na.$$






                                                share|cite|improve this answer









                                                $endgroup$
















                                                  1












                                                  1








                                                  1





                                                  $begingroup$

                                                  You are summing $n$ terms, all equal to $a$. So
                                                  $$sum_{i=0}^{n-1} a = a+ a+dotsb + a = na.$$






                                                  share|cite|improve this answer









                                                  $endgroup$



                                                  You are summing $n$ terms, all equal to $a$. So
                                                  $$sum_{i=0}^{n-1} a = a+ a+dotsb + a = na.$$







                                                  share|cite|improve this answer












                                                  share|cite|improve this answer



                                                  share|cite|improve this answer










                                                  answered yesterday









                                                  HugoHugo

                                                  8206




                                                  8206























                                                      0












                                                      $begingroup$

                                                      Hint:



                                                      $a$ times the number of terms.






                                                      share|cite|improve this answer









                                                      $endgroup$


















                                                        0












                                                        $begingroup$

                                                        Hint:



                                                        $a$ times the number of terms.






                                                        share|cite|improve this answer









                                                        $endgroup$
















                                                          0












                                                          0








                                                          0





                                                          $begingroup$

                                                          Hint:



                                                          $a$ times the number of terms.






                                                          share|cite|improve this answer









                                                          $endgroup$



                                                          Hint:



                                                          $a$ times the number of terms.







                                                          share|cite|improve this answer












                                                          share|cite|improve this answer



                                                          share|cite|improve this answer










                                                          answered yesterday









                                                          Yves DaoustYves Daoust

                                                          130k676229




                                                          130k676229






























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