Libmonster ID: U.S.-1575
Author(s) of the publication: Verner Andrei

Front side: :

 2r2.JPG

Progress Sums:-1,-2,-3,-4,-5...    1,2,3,4,5...

We express the formulas:Sn= (n²a₁+n)/2, ; Sn-1=(n²a₁-n)/2,.  (n - Number of summing members, a₁ -  first member of the progression. With a negative or positive value n. Expressions Sn-1, Sn-2  should be understood: subtraction from the number of the member taken).

First option:

   

Example: Sn= (n²a₁+n)/2.

For n = -5 we have: ((-5)2(-1)+(-5))/2=-15;                          For n = 5 we have: (52*1+5)/2=15.

 

 Example: Sn-1=(n²a₁-n)/2

 For n = -5 we have: ((-5)2(-1)-(-5))/2=-10                            For n = 5 we have: (52*1-5)/2=10.

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                          Triangular:                                                                                                                                                                                                         

 

 

 

  Progress Sums:-1,-3,-6,-10,-15....1,3,6,10,15....                                      

  We express the formulas:Sn= ((n+a₁)3-(n+a₁))/6, Sn= (n3-n)/6+(n²a₁+n)/2;  Sn-1=(n3-n)/6; Sn-2=((n-a₁)3-(n-a₁))/6,         Sn-2=(n3-n)/6-(n²a₁-n)/2.       

  First option:

 

 

  Example: Sn= ((n+a₁)3-(n+a₁))/6.

  For n = -5 we have: ((-5+(-1))3-(-5+(-1)))/6=-35;             For n = 5 we have: ((5+1)3-(5+1))/6=35.

 

  Example: Sn-2=((n-a₁)3-(n-a₁))/6.

  For n = -5 we have:((-5-(-1))3-(-5-(-1)))/6=-10;                For n = 5 we have: ((5-1)3-(5-1))/6=10.

 

  Second option:

 

  Example: Sn= (n3-n)/6+(n²a₁+n)/2.  

  For n = -5 we have: ((-5)3-(-5))/6+((-5)2(-1)+(-5))/2= -35;    For n = 5 we have:  (( 5)3-5)/6+(52*1+5)/2= 35.

          

  Example: Sn-1=(n3-n)/6.

  For n = -5 we have: ((-5)3-(-5))/6= -20;                             For n = 5 we have: (( 5)3-5)/6=20.

     

  Example: Sn-2=(n3-n)/6-(n²a₁-n)/2.

  For n = -5 we have:((-5)3-(-5))/6 -((-5)2(-1)-(-5))/2= -10;       For n = 5 we have: (( 5)3-5)/6-(52*1-5)/2= 10.

                                              

         Quadrilateral:

 4r2.JPG

 Progress Sums: -1,-4,-9,-16,-25....1,4,9,16,25....

  We express the formulas:Sn= a₁(n+a₁)(n²a₁+0,5n)/3,  Sn= (n3-n)/3 + (n²a₁+n)/2;  Sn-1= a₁(n-a₁)(n²a₁-0,5n)/3, Sn-1=(n3-n)/3 - (n²a₁-n)/2.

  First option:

 

  Example: Sn=a₁(n+a₁)(n²a₁+0,5n)/3.

  For n = -5 we have: -1(-5+(-1))*((-5)2(-1)+(-2,5))/3=-55;        For n = 5 we have: 1(5+1)(52*1+2,5)/3=55.

 

  Example: Sn-1= a₁(n-a₁)(n²a₁-0,5n)/3.

  For n = -5 we have: -1(-5-(-1))*((-5)2(-1)-(-2,5))/3=-30;           For n = 5 we have:1(5-1)(52*1-2,5)/3=30.

 

  Second option:

 

  Example: Sn= (n3-n)/3 + (n²a₁+n)/2.

  For n = -5 we have: ((-5)3-(-5))/3 +((-5)2(-1)+(-5))/2= -55;       For n = 5 we have:((5)3-5)/3 + (52*1+5)/2= 55.

 

  Example: Sn-1= (n3-n)/3 -(n²a₁-n)/2.

  For n = -5 we have: ((-5)3-(-5))/3 -((-5)2(-1)-(-5))/2= -30;           For n = 5 we have: ((5)3-5)/3 -(52*1-5)/2= 30.

 

 

 


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Verner Andrei, Calculation of the pyramid (Sequence of numerical progressions). // New-York: Libmonster (LIBMONSTER.COM). Updated: 19.07.2024. URL: https://libmonster.com/m/articles/view/Calculation-of-the-pyramid-Sequence-of-numerical-progressions (date of access: 24.07.2024).

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