Calculation of the pyramid (Sequence of numerical progressions).

[Andrei_62]

Front side:

Progress 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 - 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= (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.

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

Edited July 20, 2017 by Andrei_62

[swansont]

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Is there a question here, or a direction for the discussion?

[Acme]

Pyramids are 3-dimensional; polygonal numbers are 2-dimensional. They are not interchangeable.