Question 973153
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Triangle:  *[tex \LARGE \frac{1}{2}n(n\ +\ 1)]


Square:    *[tex \LARGE n^2]


Pentagonal:  *[tex \LARGE \frac{1}{2}n(3n\ -\ 1)]


Hexagonal:  *[tex \LARGE n(2n\ -\ 1)]


Heptagonal:  *[tex \LARGE \frac{1}{2}n(5n\ -\ 3)]


The pattern looks like octagonal should be *[tex \LARGE n(3n\ -\ 2)]


And that is indeed the correct formula.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \