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Published 2017-07-19 Published on SciPeople2017-07-19 12:49:32


Calculation of the Pyramid. (Sequence of numerical progressions)
Front side: Progressive sums:-1,-2,-3,-4,-5... 1,2,3,4,5... We express the formulas: Sn= (a₁n²+n)/2, ; Sn-1=(a₁n²-n)/2 ("n" is the number of summable terms, and "a₁" is the first term of the progression.) For negative or positive values of "n", the expressions Sn-1, Sn-2 should be understood as: subtraction from the number of the member taken). Example: Sn= (a₁n2+n)/2 For n = -5 we have: (-1*(-5)²+(-5))/2=-15; For n = 5 we have: (1*(5)²+5)/2=15 Example: Sn-1=(a₁n²-n)/2 For n = -5 we have: (-1*(-5)2-(-5))/2=-10; For n = 5 we have: (1*(5)²-5)/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+(a₁n²+n)/2; Sn-1=(n³-n)/6; Sn-2=((n-a₁)³-(n-a₁))/6, Sn-2=(n³-n)/6-(a₁n²-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+(a₁n²+n)/2. При n= -5 имеем: ((-5)³-(-5))/6+(-1*(-5)²+(-5))/2= -35; При n= 5 имеем: (( 5)³-5)/6+(1*(5)²+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-(a₁n²-n)/2. For n = -5 we have:((-5)³-(-5))/6 -(-1*(-5)²-(-5))/2= -10; For n = 5 we have: (( 5)³-5)/6-(1*(5)²-5)/2= 10. Quadrilateral: Progress Sums: -1,-4,-9,-16,-25....1,4,9,16,25.... We express the formulas: Sn= a₁(n+a₁)(a₁n²+0,5n)/3, Sn= (n³-n)/3 + (a₁n²+n)/2; Sn-1= a₁(n-a₁)(a₁n²-0,5n)/3, Sn-1=(n³-n)/3 - (a₁n²-n)/2. First option: Example: Sn=a₁(n+a₁)(a₁n²+0,5n)/3. For n = -5 we have: -1(-5+(-1))*(-1*(-5)²+(-2,5))/3=-55; For n = 5 we have: 1(5+1)(1*(5)²+2,5)/3=55. Example: Sn-1= a₁(n-a₁)(a₁n²-0,5n)/3. For n = -5 we have: -1(-5-(-1))*(-1*(-5)²-(-2,5))/3=-30; For n = 5 we have:1(5-1)(1*(5)²-2,5)/3=30. Second option: Example: Sn= (n³-n)/3 + (a₁n²+n)/2. For n = -5 we have: ((-5)³-(-5))/3 + (-1*(-5)²+(-5))/2= -55; For n = 5 we have:((5)³-5)/3 + (1*(5)²+5)/2= 55. Example: Sn-1= (n³-n)/3 -(a₁n²-n)/2. For n = -5 we have: ((-5)³-(-5))/3 -(-1*(-5)²-(-5))/2= -30; For n = 5 we have: ((5)³-5)/3 -(1*(5)²-5)/2= 30.
Abstract The sums of progressions: 1,2,3,4,5..., -1,-2,-3,-4,-5... can be found using the formula: Sn= (a₁n²+n)/2. The sums of progressions: 1,3,6,10,15..., -1,-3,-6,-10,-15... can be found using the formula:Sn= ((n+a₁)³-(n+a₁))/6. The sums of progressions: 1,4,9,16,25..., -1,-4,-9,-16,-25... can be found with the help of the formula:Sn= a₁(n+a₁)(a₁n²+0.5n)/3... (Where "n" is the number of summable terms, and "a₁" is the first member of the progression).

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