Reference Request for solution of Ramanujan Identities












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Where can I find the identities of Ramanujan concerning the Floor Function with its solution? Any site you can recomend to me? enter image description here






elementary-number-theory reference-request floor-function







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  • $begingroup$
    science.ncue.edu.tw/journal/article/1-2-7.pdf
    $endgroup$
    – Karn Watcharasupat
    Jul 28 '18 at 14:49










  • $begingroup$
    (iii) Can be found here: math.stackexchange.com/questions/6087/…
    $endgroup$
    – Robert Z
    Jul 28 '18 at 15:09
















2












$begingroup$


Where can I find the identities of Ramanujan concerning the Floor Function with its solution? Any site you can recomend to me? enter image description here






elementary-number-theory reference-request floor-function







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$endgroup$












  • $begingroup$
    science.ncue.edu.tw/journal/article/1-2-7.pdf
    $endgroup$
    – Karn Watcharasupat
    Jul 28 '18 at 14:49










  • $begingroup$
    (iii) Can be found here: math.stackexchange.com/questions/6087/…
    $endgroup$
    – Robert Z
    Jul 28 '18 at 15:09














2












2








2


1



$begingroup$


Where can I find the identities of Ramanujan concerning the Floor Function with its solution? Any site you can recomend to me? enter image description here






elementary-number-theory reference-request floor-function







share|cite|improve this question









$endgroup$




Where can I find the identities of Ramanujan concerning the Floor Function with its solution? Any site you can recomend to me? enter image description here






elementary-number-theory reference-request floor-function




elementary-number-theory reference-request floor-function






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asked Jul 28 '18 at 14:33









user573497user573497

16619




16619












  • $begingroup$
    science.ncue.edu.tw/journal/article/1-2-7.pdf
    $endgroup$
    – Karn Watcharasupat
    Jul 28 '18 at 14:49










  • $begingroup$
    (iii) Can be found here: math.stackexchange.com/questions/6087/…
    $endgroup$
    – Robert Z
    Jul 28 '18 at 15:09


















  • $begingroup$
    science.ncue.edu.tw/journal/article/1-2-7.pdf
    $endgroup$
    – Karn Watcharasupat
    Jul 28 '18 at 14:49










  • $begingroup$
    (iii) Can be found here: math.stackexchange.com/questions/6087/…
    $endgroup$
    – Robert Z
    Jul 28 '18 at 15:09
















$begingroup$
science.ncue.edu.tw/journal/article/1-2-7.pdf
$endgroup$
– Karn Watcharasupat
Jul 28 '18 at 14:49




$begingroup$
science.ncue.edu.tw/journal/article/1-2-7.pdf
$endgroup$
– Karn Watcharasupat
Jul 28 '18 at 14:49












$begingroup$
(iii) Can be found here: math.stackexchange.com/questions/6087/…
$endgroup$
– Robert Z
Jul 28 '18 at 15:09




$begingroup$
(iii) Can be found here: math.stackexchange.com/questions/6087/…
$endgroup$
– Robert Z
Jul 28 '18 at 15:09










2 Answers
2






active

oldest

votes


















2












$begingroup$

For (i), write $n=6k+r$ where $rin {0,1,...,5}$. Then left side is $$begin{eqnarray}Big[{nover 3}Big]+ Big[{n+2over 6}Big]+Big[{n+4over 6}Big] &= &Big[{6k+rover 3}Big]+ Big[{6k+r+2over 6}Big]+Big[{6k+r+4over 6}Big]\
&= &2k+Big[{rover 3}Big]+ k+Big[{r+2over 6}Big]+k+Big[{r+4over 6}Big]\
&= &4k+underbrace{Big[{rover 3}Big]+Big[{r+2over 6}Big]+Big[{r+4over 6}Big]}_{E_r}\
end{eqnarray}$$
$$E_r=left{%
begin{array}{ll}
0, & r=0,1\
1, & r=2 \
2, & r=3 \
3, & r=4,5 \
end{array}%
right.$$
And the right side is
$$begin{eqnarray}Big[{nover 2}Big]+ Big[{n+3over 6}Big]&= &Big[{6k+rover 2}Big]+ Big[{6k+r+3over 6}Big]\
&= &3k+Big[{rover 2}Big]+ k+Big[{r+3over 6}Big]\
&= &4k+underbrace{Big[{rover 2}Big]+Big[{r+3over 6}Big]}_{F_r}\
end{eqnarray}$$
$$F_r=left{%
begin{array}{ll}
0, & r=0,1\
1, & r=2 \
2, & r=3 \
3, & r=4,5 \
end{array}%
right.$$
So both sides are the same for all $r$ and we are done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why did you just let $n=6k+r$? Is this the general form of an positive integer?
    $endgroup$
    – user573497
    Jul 28 '18 at 15:56










  • $begingroup$
    That is correct. If we had $[n/11]$ then I would write $n=11k+r$ and $0leq rleq 10$
    $endgroup$
    – greedoid
    Jul 28 '18 at 16:08






  • 1




    $begingroup$
    oh, I get it! You did that for convenience since in the denominator there is a $6$. If it were some arbitrary number un the denominator like $m$, then I consider the integer $n=mk+r$ where 0leqrls
    $endgroup$
    – user573497
    Jul 28 '18 at 16:33






  • 1




    $begingroup$
    Sorry, I couldn't edit the last comment. I tried to say that $n=mk+r$ where $r$ is an integer between $0$ and $m-1$.
    $endgroup$
    – user573497
    Jul 28 '18 at 16:36










  • $begingroup$
    I can understand now! Thanks for the feedback! :)
    $endgroup$
    – user573497
    Jul 28 '18 at 16:38



















0












$begingroup$

If $[x]$ denotes floor we have
$$sumlimits_{k=0}^{m-1}Big[{n+ks+tover ms}Big]=Big[{n+tover s}Big]$$
for $ngeqslant0$, $m>0$, $s>tgeqslant0$, $n,m,s,t$ - integers. So
$$Big[{nover 2}Big]=Big[{nover 6}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]$$
$$Big[{nover 3}Big]=Big[{nover 6}Big]+Big[{n+3over 6}Big]$$
and obviously
$$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{nover 2}Big]+Big[{n+3over 6}Big]$$
With same conditions for $n,m,s,t$ if $[x]$ denotes ceiling we have
$$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$
and also if $[x]$ denotes nearest integer we have
$$Big[{n+2over 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

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    2












    $begingroup$

    For (i), write $n=6k+r$ where $rin {0,1,...,5}$. Then left side is $$begin{eqnarray}Big[{nover 3}Big]+ Big[{n+2over 6}Big]+Big[{n+4over 6}Big] &= &Big[{6k+rover 3}Big]+ Big[{6k+r+2over 6}Big]+Big[{6k+r+4over 6}Big]\
    &= &2k+Big[{rover 3}Big]+ k+Big[{r+2over 6}Big]+k+Big[{r+4over 6}Big]\
    &= &4k+underbrace{Big[{rover 3}Big]+Big[{r+2over 6}Big]+Big[{r+4over 6}Big]}_{E_r}\
    end{eqnarray}$$
    $$E_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    And the right side is
    $$begin{eqnarray}Big[{nover 2}Big]+ Big[{n+3over 6}Big]&= &Big[{6k+rover 2}Big]+ Big[{6k+r+3over 6}Big]\
    &= &3k+Big[{rover 2}Big]+ k+Big[{r+3over 6}Big]\
    &= &4k+underbrace{Big[{rover 2}Big]+Big[{r+3over 6}Big]}_{F_r}\
    end{eqnarray}$$
    $$F_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    So both sides are the same for all $r$ and we are done.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why did you just let $n=6k+r$? Is this the general form of an positive integer?
      $endgroup$
      – user573497
      Jul 28 '18 at 15:56










    • $begingroup$
      That is correct. If we had $[n/11]$ then I would write $n=11k+r$ and $0leq rleq 10$
      $endgroup$
      – greedoid
      Jul 28 '18 at 16:08






    • 1




      $begingroup$
      oh, I get it! You did that for convenience since in the denominator there is a $6$. If it were some arbitrary number un the denominator like $m$, then I consider the integer $n=mk+r$ where 0leqrls
      $endgroup$
      – user573497
      Jul 28 '18 at 16:33






    • 1




      $begingroup$
      Sorry, I couldn't edit the last comment. I tried to say that $n=mk+r$ where $r$ is an integer between $0$ and $m-1$.
      $endgroup$
      – user573497
      Jul 28 '18 at 16:36










    • $begingroup$
      I can understand now! Thanks for the feedback! :)
      $endgroup$
      – user573497
      Jul 28 '18 at 16:38
















    2












    $begingroup$

    For (i), write $n=6k+r$ where $rin {0,1,...,5}$. Then left side is $$begin{eqnarray}Big[{nover 3}Big]+ Big[{n+2over 6}Big]+Big[{n+4over 6}Big] &= &Big[{6k+rover 3}Big]+ Big[{6k+r+2over 6}Big]+Big[{6k+r+4over 6}Big]\
    &= &2k+Big[{rover 3}Big]+ k+Big[{r+2over 6}Big]+k+Big[{r+4over 6}Big]\
    &= &4k+underbrace{Big[{rover 3}Big]+Big[{r+2over 6}Big]+Big[{r+4over 6}Big]}_{E_r}\
    end{eqnarray}$$
    $$E_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    And the right side is
    $$begin{eqnarray}Big[{nover 2}Big]+ Big[{n+3over 6}Big]&= &Big[{6k+rover 2}Big]+ Big[{6k+r+3over 6}Big]\
    &= &3k+Big[{rover 2}Big]+ k+Big[{r+3over 6}Big]\
    &= &4k+underbrace{Big[{rover 2}Big]+Big[{r+3over 6}Big]}_{F_r}\
    end{eqnarray}$$
    $$F_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    So both sides are the same for all $r$ and we are done.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why did you just let $n=6k+r$? Is this the general form of an positive integer?
      $endgroup$
      – user573497
      Jul 28 '18 at 15:56










    • $begingroup$
      That is correct. If we had $[n/11]$ then I would write $n=11k+r$ and $0leq rleq 10$
      $endgroup$
      – greedoid
      Jul 28 '18 at 16:08






    • 1




      $begingroup$
      oh, I get it! You did that for convenience since in the denominator there is a $6$. If it were some arbitrary number un the denominator like $m$, then I consider the integer $n=mk+r$ where 0leqrls
      $endgroup$
      – user573497
      Jul 28 '18 at 16:33






    • 1




      $begingroup$
      Sorry, I couldn't edit the last comment. I tried to say that $n=mk+r$ where $r$ is an integer between $0$ and $m-1$.
      $endgroup$
      – user573497
      Jul 28 '18 at 16:36










    • $begingroup$
      I can understand now! Thanks for the feedback! :)
      $endgroup$
      – user573497
      Jul 28 '18 at 16:38














    2












    2








    2





    $begingroup$

    For (i), write $n=6k+r$ where $rin {0,1,...,5}$. Then left side is $$begin{eqnarray}Big[{nover 3}Big]+ Big[{n+2over 6}Big]+Big[{n+4over 6}Big] &= &Big[{6k+rover 3}Big]+ Big[{6k+r+2over 6}Big]+Big[{6k+r+4over 6}Big]\
    &= &2k+Big[{rover 3}Big]+ k+Big[{r+2over 6}Big]+k+Big[{r+4over 6}Big]\
    &= &4k+underbrace{Big[{rover 3}Big]+Big[{r+2over 6}Big]+Big[{r+4over 6}Big]}_{E_r}\
    end{eqnarray}$$
    $$E_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    And the right side is
    $$begin{eqnarray}Big[{nover 2}Big]+ Big[{n+3over 6}Big]&= &Big[{6k+rover 2}Big]+ Big[{6k+r+3over 6}Big]\
    &= &3k+Big[{rover 2}Big]+ k+Big[{r+3over 6}Big]\
    &= &4k+underbrace{Big[{rover 2}Big]+Big[{r+3over 6}Big]}_{F_r}\
    end{eqnarray}$$
    $$F_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    So both sides are the same for all $r$ and we are done.






    share|cite|improve this answer









    $endgroup$



    For (i), write $n=6k+r$ where $rin {0,1,...,5}$. Then left side is $$begin{eqnarray}Big[{nover 3}Big]+ Big[{n+2over 6}Big]+Big[{n+4over 6}Big] &= &Big[{6k+rover 3}Big]+ Big[{6k+r+2over 6}Big]+Big[{6k+r+4over 6}Big]\
    &= &2k+Big[{rover 3}Big]+ k+Big[{r+2over 6}Big]+k+Big[{r+4over 6}Big]\
    &= &4k+underbrace{Big[{rover 3}Big]+Big[{r+2over 6}Big]+Big[{r+4over 6}Big]}_{E_r}\
    end{eqnarray}$$
    $$E_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    And the right side is
    $$begin{eqnarray}Big[{nover 2}Big]+ Big[{n+3over 6}Big]&= &Big[{6k+rover 2}Big]+ Big[{6k+r+3over 6}Big]\
    &= &3k+Big[{rover 2}Big]+ k+Big[{r+3over 6}Big]\
    &= &4k+underbrace{Big[{rover 2}Big]+Big[{r+3over 6}Big]}_{F_r}\
    end{eqnarray}$$
    $$F_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    So both sides are the same for all $r$ and we are done.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jul 28 '18 at 14:51









    greedoidgreedoid

    40.1k114799




    40.1k114799












    • $begingroup$
      Why did you just let $n=6k+r$? Is this the general form of an positive integer?
      $endgroup$
      – user573497
      Jul 28 '18 at 15:56










    • $begingroup$
      That is correct. If we had $[n/11]$ then I would write $n=11k+r$ and $0leq rleq 10$
      $endgroup$
      – greedoid
      Jul 28 '18 at 16:08






    • 1




      $begingroup$
      oh, I get it! You did that for convenience since in the denominator there is a $6$. If it were some arbitrary number un the denominator like $m$, then I consider the integer $n=mk+r$ where 0leqrls
      $endgroup$
      – user573497
      Jul 28 '18 at 16:33






    • 1




      $begingroup$
      Sorry, I couldn't edit the last comment. I tried to say that $n=mk+r$ where $r$ is an integer between $0$ and $m-1$.
      $endgroup$
      – user573497
      Jul 28 '18 at 16:36










    • $begingroup$
      I can understand now! Thanks for the feedback! :)
      $endgroup$
      – user573497
      Jul 28 '18 at 16:38


















    • $begingroup$
      Why did you just let $n=6k+r$? Is this the general form of an positive integer?
      $endgroup$
      – user573497
      Jul 28 '18 at 15:56










    • $begingroup$
      That is correct. If we had $[n/11]$ then I would write $n=11k+r$ and $0leq rleq 10$
      $endgroup$
      – greedoid
      Jul 28 '18 at 16:08






    • 1




      $begingroup$
      oh, I get it! You did that for convenience since in the denominator there is a $6$. If it were some arbitrary number un the denominator like $m$, then I consider the integer $n=mk+r$ where 0leqrls
      $endgroup$
      – user573497
      Jul 28 '18 at 16:33






    • 1




      $begingroup$
      Sorry, I couldn't edit the last comment. I tried to say that $n=mk+r$ where $r$ is an integer between $0$ and $m-1$.
      $endgroup$
      – user573497
      Jul 28 '18 at 16:36










    • $begingroup$
      I can understand now! Thanks for the feedback! :)
      $endgroup$
      – user573497
      Jul 28 '18 at 16:38
















    $begingroup$
    Why did you just let $n=6k+r$? Is this the general form of an positive integer?
    $endgroup$
    – user573497
    Jul 28 '18 at 15:56




    $begingroup$
    Why did you just let $n=6k+r$? Is this the general form of an positive integer?
    $endgroup$
    – user573497
    Jul 28 '18 at 15:56












    $begingroup$
    That is correct. If we had $[n/11]$ then I would write $n=11k+r$ and $0leq rleq 10$
    $endgroup$
    – greedoid
    Jul 28 '18 at 16:08




    $begingroup$
    That is correct. If we had $[n/11]$ then I would write $n=11k+r$ and $0leq rleq 10$
    $endgroup$
    – greedoid
    Jul 28 '18 at 16:08




    1




    1




    $begingroup$
    oh, I get it! You did that for convenience since in the denominator there is a $6$. If it were some arbitrary number un the denominator like $m$, then I consider the integer $n=mk+r$ where 0leqrls
    $endgroup$
    – user573497
    Jul 28 '18 at 16:33




    $begingroup$
    oh, I get it! You did that for convenience since in the denominator there is a $6$. If it were some arbitrary number un the denominator like $m$, then I consider the integer $n=mk+r$ where 0leqrls
    $endgroup$
    – user573497
    Jul 28 '18 at 16:33




    1




    1




    $begingroup$
    Sorry, I couldn't edit the last comment. I tried to say that $n=mk+r$ where $r$ is an integer between $0$ and $m-1$.
    $endgroup$
    – user573497
    Jul 28 '18 at 16:36




    $begingroup$
    Sorry, I couldn't edit the last comment. I tried to say that $n=mk+r$ where $r$ is an integer between $0$ and $m-1$.
    $endgroup$
    – user573497
    Jul 28 '18 at 16:36












    $begingroup$
    I can understand now! Thanks for the feedback! :)
    $endgroup$
    – user573497
    Jul 28 '18 at 16:38




    $begingroup$
    I can understand now! Thanks for the feedback! :)
    $endgroup$
    – user573497
    Jul 28 '18 at 16:38











    0












    $begingroup$

    If $[x]$ denotes floor we have
    $$sumlimits_{k=0}^{m-1}Big[{n+ks+tover ms}Big]=Big[{n+tover s}Big]$$
    for $ngeqslant0$, $m>0$, $s>tgeqslant0$, $n,m,s,t$ - integers. So
    $$Big[{nover 2}Big]=Big[{nover 6}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]$$
    $$Big[{nover 3}Big]=Big[{nover 6}Big]+Big[{n+3over 6}Big]$$
    and obviously
    $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{nover 2}Big]+Big[{n+3over 6}Big]$$
    With same conditions for $n,m,s,t$ if $[x]$ denotes ceiling we have
    $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$
    and also if $[x]$ denotes nearest integer we have
    $$Big[{n+2over 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      If $[x]$ denotes floor we have
      $$sumlimits_{k=0}^{m-1}Big[{n+ks+tover ms}Big]=Big[{n+tover s}Big]$$
      for $ngeqslant0$, $m>0$, $s>tgeqslant0$, $n,m,s,t$ - integers. So
      $$Big[{nover 2}Big]=Big[{nover 6}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]$$
      $$Big[{nover 3}Big]=Big[{nover 6}Big]+Big[{n+3over 6}Big]$$
      and obviously
      $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{nover 2}Big]+Big[{n+3over 6}Big]$$
      With same conditions for $n,m,s,t$ if $[x]$ denotes ceiling we have
      $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$
      and also if $[x]$ denotes nearest integer we have
      $$Big[{n+2over 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        If $[x]$ denotes floor we have
        $$sumlimits_{k=0}^{m-1}Big[{n+ks+tover ms}Big]=Big[{n+tover s}Big]$$
        for $ngeqslant0$, $m>0$, $s>tgeqslant0$, $n,m,s,t$ - integers. So
        $$Big[{nover 2}Big]=Big[{nover 6}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]$$
        $$Big[{nover 3}Big]=Big[{nover 6}Big]+Big[{n+3over 6}Big]$$
        and obviously
        $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{nover 2}Big]+Big[{n+3over 6}Big]$$
        With same conditions for $n,m,s,t$ if $[x]$ denotes ceiling we have
        $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$
        and also if $[x]$ denotes nearest integer we have
        $$Big[{n+2over 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$






        share|cite|improve this answer











        $endgroup$



        If $[x]$ denotes floor we have
        $$sumlimits_{k=0}^{m-1}Big[{n+ks+tover ms}Big]=Big[{n+tover s}Big]$$
        for $ngeqslant0$, $m>0$, $s>tgeqslant0$, $n,m,s,t$ - integers. So
        $$Big[{nover 2}Big]=Big[{nover 6}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]$$
        $$Big[{nover 3}Big]=Big[{nover 6}Big]+Big[{n+3over 6}Big]$$
        and obviously
        $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{nover 2}Big]+Big[{n+3over 6}Big]$$
        With same conditions for $n,m,s,t$ if $[x]$ denotes ceiling we have
        $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$
        and also if $[x]$ denotes nearest integer we have
        $$Big[{n+2over 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 2 at 18:42

























        answered Jan 2 at 18:15









        user514787user514787

        715210




        715210






























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