Compute $S = sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor$












3












$begingroup$


I want to compute the following sum
$$S = sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor.$$
Here is what I tried:
$$ S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor.$$
If $m= 2t$ then
$$S =sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{(t-1)t}{2} = t^2.$$
If $m= 2t+1$ then
$$S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{t(t+1)}{2}= t(t+1).$$



But I am not sure if this is correct. Perhaps someone could give an indication.






sequences-and-series number-theory







share|cite|improve this question











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  • 1




    $begingroup$
    You can write $t$ in terms of $m$ in each case.
    $endgroup$
    – Hello_World
    2 days ago






  • 1




    $begingroup$
    @Hello_World Actuallly, you can.
    $endgroup$
    – 5xum
    2 days ago










  • $begingroup$
    Are you asking me to write them in terms of $m$? I can do that if it helps.
    $endgroup$
    – Hello_World
    2 days ago












  • $begingroup$
    @KemonoChen, please write that as an answer so that we can downvote it.
    $endgroup$
    – Carsten S
    2 days ago
















3












$begingroup$


I want to compute the following sum
$$S = sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor.$$
Here is what I tried:
$$ S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor.$$
If $m= 2t$ then
$$S =sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{(t-1)t}{2} = t^2.$$
If $m= 2t+1$ then
$$S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{t(t+1)}{2}= t(t+1).$$



But I am not sure if this is correct. Perhaps someone could give an indication.






sequences-and-series number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You can write $t$ in terms of $m$ in each case.
    $endgroup$
    – Hello_World
    2 days ago






  • 1




    $begingroup$
    @Hello_World Actuallly, you can.
    $endgroup$
    – 5xum
    2 days ago










  • $begingroup$
    Are you asking me to write them in terms of $m$? I can do that if it helps.
    $endgroup$
    – Hello_World
    2 days ago












  • $begingroup$
    @KemonoChen, please write that as an answer so that we can downvote it.
    $endgroup$
    – Carsten S
    2 days ago














3












3








3





$begingroup$


I want to compute the following sum
$$S = sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor.$$
Here is what I tried:
$$ S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor.$$
If $m= 2t$ then
$$S =sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{(t-1)t}{2} = t^2.$$
If $m= 2t+1$ then
$$S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{t(t+1)}{2}= t(t+1).$$



But I am not sure if this is correct. Perhaps someone could give an indication.






sequences-and-series number-theory







share|cite|improve this question











$endgroup$




I want to compute the following sum
$$S = sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor.$$
Here is what I tried:
$$ S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor.$$
If $m= 2t$ then
$$S =sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{(t-1)t}{2} = t^2.$$
If $m= 2t+1$ then
$$S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{t(t+1)}{2}= t(t+1).$$



But I am not sure if this is correct. Perhaps someone could give an indication.






sequences-and-series number-theory




sequences-and-series number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Asaf Karagila♦

302k32427757




302k32427757










asked 2 days ago









Hello_WorldHello_World

4,16221631




4,16221631








  • 1




    $begingroup$
    You can write $t$ in terms of $m$ in each case.
    $endgroup$
    – Hello_World
    2 days ago






  • 1




    $begingroup$
    @Hello_World Actuallly, you can.
    $endgroup$
    – 5xum
    2 days ago










  • $begingroup$
    Are you asking me to write them in terms of $m$? I can do that if it helps.
    $endgroup$
    – Hello_World
    2 days ago












  • $begingroup$
    @KemonoChen, please write that as an answer so that we can downvote it.
    $endgroup$
    – Carsten S
    2 days ago














  • 1




    $begingroup$
    You can write $t$ in terms of $m$ in each case.
    $endgroup$
    – Hello_World
    2 days ago






  • 1




    $begingroup$
    @Hello_World Actuallly, you can.
    $endgroup$
    – 5xum
    2 days ago










  • $begingroup$
    Are you asking me to write them in terms of $m$? I can do that if it helps.
    $endgroup$
    – Hello_World
    2 days ago












  • $begingroup$
    @KemonoChen, please write that as an answer so that we can downvote it.
    $endgroup$
    – Carsten S
    2 days ago








1




1




$begingroup$
You can write $t$ in terms of $m$ in each case.
$endgroup$
– Hello_World
2 days ago




$begingroup$
You can write $t$ in terms of $m$ in each case.
$endgroup$
– Hello_World
2 days ago




1




1




$begingroup$
@Hello_World Actuallly, you can.
$endgroup$
– 5xum
2 days ago




$begingroup$
@Hello_World Actuallly, you can.
$endgroup$
– 5xum
2 days ago












$begingroup$
Are you asking me to write them in terms of $m$? I can do that if it helps.
$endgroup$
– Hello_World
2 days ago






$begingroup$
Are you asking me to write them in terms of $m$? I can do that if it helps.
$endgroup$
– Hello_World
2 days ago














$begingroup$
@KemonoChen, please write that as an answer so that we can downvote it.
$endgroup$
– Carsten S
2 days ago




$begingroup$
@KemonoChen, please write that as an answer so that we can downvote it.
$endgroup$
– Carsten S
2 days ago










1 Answer
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8












$begingroup$

Yes, you are correct. You may also write the result as a more compact formula:
$$sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor=
begin{cases}
t^2&text {if $m=2t$}\
t(t+1)&text {if $m=2t+1$}\
end{cases}=leftlfloor frac{m^2}{4}rightrfloor.$$
Indeed, if $m=2t$ then
$$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2rightrfloor=t^2$$
and if $m=2t+1$ then
$$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2+t+frac{1}{4}rightrfloor=t(t+1).$$






share|cite|improve this answer











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    8












    $begingroup$

    Yes, you are correct. You may also write the result as a more compact formula:
    $$sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor=
    begin{cases}
    t^2&text {if $m=2t$}\
    t(t+1)&text {if $m=2t+1$}\
    end{cases}=leftlfloor frac{m^2}{4}rightrfloor.$$
    Indeed, if $m=2t$ then
    $$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2rightrfloor=t^2$$
    and if $m=2t+1$ then
    $$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2+t+frac{1}{4}rightrfloor=t(t+1).$$






    share|cite|improve this answer











    $endgroup$


















      8












      $begingroup$

      Yes, you are correct. You may also write the result as a more compact formula:
      $$sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor=
      begin{cases}
      t^2&text {if $m=2t$}\
      t(t+1)&text {if $m=2t+1$}\
      end{cases}=leftlfloor frac{m^2}{4}rightrfloor.$$
      Indeed, if $m=2t$ then
      $$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2rightrfloor=t^2$$
      and if $m=2t+1$ then
      $$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2+t+frac{1}{4}rightrfloor=t(t+1).$$






      share|cite|improve this answer











      $endgroup$
















        8












        8








        8





        $begingroup$

        Yes, you are correct. You may also write the result as a more compact formula:
        $$sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor=
        begin{cases}
        t^2&text {if $m=2t$}\
        t(t+1)&text {if $m=2t+1$}\
        end{cases}=leftlfloor frac{m^2}{4}rightrfloor.$$
        Indeed, if $m=2t$ then
        $$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2rightrfloor=t^2$$
        and if $m=2t+1$ then
        $$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2+t+frac{1}{4}rightrfloor=t(t+1).$$






        share|cite|improve this answer











        $endgroup$



        Yes, you are correct. You may also write the result as a more compact formula:
        $$sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor=
        begin{cases}
        t^2&text {if $m=2t$}\
        t(t+1)&text {if $m=2t+1$}\
        end{cases}=leftlfloor frac{m^2}{4}rightrfloor.$$
        Indeed, if $m=2t$ then
        $$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2rightrfloor=t^2$$
        and if $m=2t+1$ then
        $$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2+t+frac{1}{4}rightrfloor=t(t+1).$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered 2 days ago









        Robert ZRobert Z

        94.5k1063134




        94.5k1063134






























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