Mathematics Dictionary

Mathematics Dictionary
Dr. K. G. Shih

Triangular Number

Questions

Symbol Defintion

Sqr(x) = Square root of x

Q01 | - Triangular number in Pascal triangle

Q02 | - Triangular number : Patterns

Q03 | - Triangular number : Prove T(n) = n*(n+1)/2

Q04 | - Triangular number : Prove Sum[T(n)] = n*(n+1)*(n+2)/6

Q05 | - Triangular number : Prove that Sum[C(n+1),2)] = C(n+2,3)

Q06 | - Using Sum[n*(n+1)/2] find Sum[n^2] = n*(n+1)*(2*n+1)/6

Q07 | -

Q08 | -

Q09 | -

Q10 | -

Answers

Q01. Triangular 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

Triangular number

  1. Triandular numbers in column 3 (C3)
  2. The numbers : 1, 3, 6, 10, 15, 21, 28, ...
  3. 1st Diff    :    2, 3,  4,  5,  6,  7, ...
  4. 2nd diff    :       1,  1,  1,  1,  1, ...

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Q02.Triangular number : Patterns
Pattern

    1    2     6      10 

    *    *     *       *
        * *   * *     * *
             * * *   * * *
                    * * * *

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Q03. Triangular number : Find T(n) = n*(n+1)/2

T(n) = n*(n+1)/2

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

Go to Begin

Q04. Triangular number : Sum[T(n)] = n*(n+1)*(n+2)/6

Proof

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

Go to Begin

Q05. Triangular number : Sum[C(n+1,2)] = C(n+2,3)

   1. Binomial expansion coefficients
      C(m,r) = m*(m-1)*(m-2) .... (m-r+1)/r!
   2. Hence (n+1)*n/2 = C(n+1,2)
            (n+2)*(n+1)*n/6 = C(n+2,3)
   3. From Q04 we have Sum[C(n+1,2)] = C(n+2,3)

Go to Begin

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

Proof

  Sum[n*(n+1)/2] = (Sum[n] + Sum[n^2])/2 ............ (1)
  Since Sum[n] = n*(n+1)/2 .......................... (2)
        Sum[n*(n+1)/2] = n*(n+1)*(n+2)/6 ............ (3)
  From (1) Sum[n^2] = 2*Sum[n*(n+1)/2] - Sum[n] ..... (4)
  Substitute (2) and (3) into (4) and simplify we have

  Sum[n^2] = 2*n*(n+1)*(n+2)/6 - n*(n+1)/2
           = n*(n+1)*(2*n+4-3)/6
           = n*(n+1)*(2*n+1)/6

Reference

  More method to prove this formula are given in keyword series
  about Sum[n^2] in www.b192907.com

Q07. Answer

Q08. Answer

Q09. Answer

Q10. Answer

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