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A formula


aommaster

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Ok this sequence starts with one triangle (a), and ends with 7 on the end (n):

 

So applying Sn. For Arithmetic Progressions:

 

x(a+n)/2

 

A = first term 1, n = last sum (7), x =number of sequence lines (4)

 

Sn. = 4(1+7)/2 = 16 (remember only the internal sequence triangles).

 

Theorise: calculating 27 lines:

 

1, 3, 5, 7 (triangles going up at rate of 2 added to previous).

 

Sp. = z -1 x d + 1

Z = Line number to find, m = constant sequence difference (2):

 

27 – 1 x d + 1 = 53

 

So the 27-sequence line would contain 53 triangles (internal).

 

So then apply Sn.

 

x(a+n)/2

 

A = first term 1, n = last sum (53), x =number of sequence lines (27)

 

27(1+53)/2 = 729.

 

The total amount of internal triangle in 27 lines would be 729.

 

I hope this helps.

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