How many triangles has a complete graph. Recursive Equation












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I need to find the solution with a recursive equation. I found the solution with combinations:
$${n}choose{}3$$



but I don't know how to find it with recursive equations. I don't find a pattern to resolve it



Thanks










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    0












    $begingroup$


    I need to find the solution with a recursive equation. I found the solution with combinations:
    $${n}choose{}3$$



    but I don't know how to find it with recursive equations. I don't find a pattern to resolve it



    Thanks










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I need to find the solution with a recursive equation. I found the solution with combinations:
      $${n}choose{}3$$



      but I don't know how to find it with recursive equations. I don't find a pattern to resolve it



      Thanks










      share|cite|improve this question









      $endgroup$




      I need to find the solution with a recursive equation. I found the solution with combinations:
      $${n}choose{}3$$



      but I don't know how to find it with recursive equations. I don't find a pattern to resolve it



      Thanks







      discrete-mathematics graph-theory combinations






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      asked Dec 6 '18 at 20:36









      emeeemee

      43




      43






















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          $begingroup$

          Let $T_n$ be the number of triangles in a complete graph of size $n$ for $ngeq 3$. Deduce that $T_{n+1} = T_n + binom{n}{2}$, as it amounts to considering if a triangle is contained in the complete subgraph on $n$ vertices or not. With $T_3=1$, it follows that $T_n=1+sum_{k=3}^{n-1}binom{k}{2}=1+sum_{k=3}^{n-1}k(k+1)/2$. Simplifying this will give you the answer.






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            $begingroup$

            Let $T_n$ be the number of triangles in a complete graph of size $n$ for $ngeq 3$. Deduce that $T_{n+1} = T_n + binom{n}{2}$, as it amounts to considering if a triangle is contained in the complete subgraph on $n$ vertices or not. With $T_3=1$, it follows that $T_n=1+sum_{k=3}^{n-1}binom{k}{2}=1+sum_{k=3}^{n-1}k(k+1)/2$. Simplifying this will give you the answer.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Let $T_n$ be the number of triangles in a complete graph of size $n$ for $ngeq 3$. Deduce that $T_{n+1} = T_n + binom{n}{2}$, as it amounts to considering if a triangle is contained in the complete subgraph on $n$ vertices or not. With $T_3=1$, it follows that $T_n=1+sum_{k=3}^{n-1}binom{k}{2}=1+sum_{k=3}^{n-1}k(k+1)/2$. Simplifying this will give you the answer.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Let $T_n$ be the number of triangles in a complete graph of size $n$ for $ngeq 3$. Deduce that $T_{n+1} = T_n + binom{n}{2}$, as it amounts to considering if a triangle is contained in the complete subgraph on $n$ vertices or not. With $T_3=1$, it follows that $T_n=1+sum_{k=3}^{n-1}binom{k}{2}=1+sum_{k=3}^{n-1}k(k+1)/2$. Simplifying this will give you the answer.






                share|cite|improve this answer









                $endgroup$



                Let $T_n$ be the number of triangles in a complete graph of size $n$ for $ngeq 3$. Deduce that $T_{n+1} = T_n + binom{n}{2}$, as it amounts to considering if a triangle is contained in the complete subgraph on $n$ vertices or not. With $T_3=1$, it follows that $T_n=1+sum_{k=3}^{n-1}binom{k}{2}=1+sum_{k=3}^{n-1}k(k+1)/2$. Simplifying this will give you the answer.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 6 '18 at 21:42









                Alex R.Alex R.

                24.9k12452




                24.9k12452






























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