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Mathematics Dictionary
Dr. K. G. Shih
Tetrahedral Numbers
Questions


  • Q01 | - Tetrahedral number in Pascal triangle
  • Q02 | - Tetrahedral number : Patterns
  • Q03 | - Tetrahedral number : nth term = T(n) = n*(n+1)*(n+2)/6
  • Q04 | - Tetrahedral number : Prove S(n) = n*(n+1)*(n+2)*(n+3)/4!
  • Q05 | - Tetrahedral number : Prove that Sum[C(n+2),3)] = C(n+3,4)
  • Q06 | - Using Sum[n*(n+1)*(n+2)/6] find Sum[n^3]
  • Q07 | - Tetrahedral number : Pattern
  • Q08 | -
  • Q09 | -
  • Q10 | -
    • Answers


    Q01. Tetrahedral number in Pascal triangle
    Pascal triangle 
          C1 C2 C3 C4 C5 C6 C7 C8 ....

  R1  1
  R2  1  1
  R3  1  2  1
  R4  1  3  3  1
  R5  1  4  6  4  1
  R6  1  5 10 10  5  1
  R7  1  6 15 20 15  6  1
  R8  1  7 21 35 35 21  7  1
  R9  1  8 28 56 70 56 28  8  1
    Tetrahedral number 
     1. Tetrahedral numbers in column 4 (C4)
 2. The numbers : 1, 4,10, 20, 35, 56, 84, ...
 3. 1st Diff    : 1, 3, 6, 10, 15, 21, 28, ...
 4. 2nd Diff    : 1, 2, 3,  4,  5,  6,  7, ...
 5. 3rd Diff    : 1, 1, 1,  1,  1,  1,  1, ...

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    Q02. Find Tetrahedral number sequence using triangular numbers
    Method 
      Tetraheral sequence : 1
  Triangular sequence : 1   3   6   10  15   21   28 .....

  2nd number of tetrahedral number is  1 +  3 = 4
  3rd number of tetrahedral number is  4 +  6 = 10
  4th number of tetrahedral number is 10 + 10 = 20
  5th number of tetrahedral number is 20 + 15 = 35
  2nd number of tetrahedral number is 35 + 21 = 56
                                       |    |    |
                                       |    |    Tetrahedral seq 
                                       |    Triangular numbers
                                       Tetrahedral seq

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    Q03. Tetrahedral number : nth term = T(n) = n*(n+1)*(n+2)/6

    T(n) = n*(n+1)/2 
       T(1) =  1
   T(2) =  4 = 1 + 3
   T(3) = 10 = 1 + 3 + 6
   T(4) = 20 = 1 + 3 + 6 + 10
   T(5) = 35 = 1 + 3 + 6 + 10 + 15
   ....
   T(n) = Sum[n*(n+1)/2] = n*(n+1)*(n+2)/6

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    Q04. Tetrahedral number : Sum[T(n)] = n*(n+1)*(n+2)*(n+3)/4!

    Proof 
       S(n) = n*(n+1)/2 ........................... (1)
   S(n^2) = n*(n+1)*(2*n+1)/6 ................. (2)
   S(n^3) = (n*(n+1)/2)^2 ..................... (3)
   Sum[T(n)] = Sum[n*(n+1)*(n+2)/6]
             = ((Sum[n^3 + 3*n^2 + 2*n)] ...... (4)
   Substitute (1), (2) and (3) into (4) and simplify we have
   Sum[T(n)] = Sum[n*(n+1)*(n+2)/6
             = n*(n+1)*(n+2)*(n+3)/4!

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    Q05. Tetrahedral : Sum[C(n+2,3)] = C(n+3,4)

       1. Binomial expansion coefficients
      C(m,r) = m*(m-1)*(m-2) .... (m-r+1)/r!
   2. Hence C(n+2,3) = (n+2)*(n+1)*n/3!
            C(n+3,4) = (n+3)*(n+2)*(n+1)*n/4!
   3. Hence from Q04 we have Sum[C(n+2,3)] = C(n+3,4)

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    Q06. Using Sum[n*(n+1)*(n+2)/6] find Sum[n^3]

    Proof : We need following formula 
      1. Formula required
     Sum[n] = n*(n+1)/2
     Sum[n^2] = n*(n+1)*(n+2)/6 
     Sum[n*(n+1)*(n+2)/6] = n*(n+1)*(n+2)*(n+3)/4!
  2. method
     Sum[n*(n+1)*(n+2)/6] = Sum[n^3 + 3*n^2 +2*n)/6]
     Sum[n^3] = 6*n*(n+1)*(n+2)*(n+3)/24 - 3*sum[n^2] - 2*sum(n)
              = ....
Reference

      More methods to prove this formula are given in keyword series
  about Sum[n^3] in www.b192907.com

    


    
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Q07. Tetrahedral number : Pattern
    

The pattern in tetrhedron (4 surfaces, each is triangle)

        *    *     * ----- 1 = top          * --------- 1 = top

         *     *                        *
        * *   * * ---- 3 = middle      * * -------- 2nd layer

               *                        *
              * *                      * *
             * * * --- 6 = bottom     * * * ------- 3rd layer

                                        *
                                       * *
                                      * * *
                                     * * * * ------ 4th layer = base

    1st number =  1
    2nd number =  4 = 1+3
    3rd number = 10 = 1+3+6 
    4th number = 20 = 1+3+6+10

   

    


    
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Q08. Answer
    

      

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Q09. Answer
    

      

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Q10. Answer
    

      

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