- #1

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The answer is n(n-3)/2.

I don't know how to do that.

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- #1

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The answer is n(n-3)/2.

I don't know how to do that.

- #2

Tide

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Set up a recursion for the number of diagonals in an n-gon: [itex]D_{n+1} = D_n + n - 2[/itex].

- #3

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I don't understand how can I set up

[itex]D_{n+1} = D_n + n - 2[/itex].

[itex]D_{n+1} = D_n + n - 2[/itex].

- #4

HallsofIvy

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