Let Tn be the number of all possible triangles formed by joining verti...
We can approach this problem by counting the number of triangles formed by selecting three vertices from the polygon.
For an n-sided polygon, there are n ways to select the first vertex, n-1 ways to select the second vertex (since we can't select the same vertex twice), and n-2 ways to select the third vertex. However, we've overcounted, since selecting vertices A, B, C is the same as selecting vertices B, C, A or C, A, B. Specifically, each triangle has been counted 3 times, so we need to divide by 3:
Tn = (n)(n-1)(n-2)/3
For Tn+1, we can use the same logic. There are n+1 ways to select the first vertex, n ways to select the second vertex, and n-1 ways to select the third vertex. Again, we've overcounted by a factor of 3:
Tn+1 = (n+1)(n)(n-1)/3
We want to find Tn+1/Tn:
(Tn+1)/(Tn) = [(n+1)(n)(n-1)/3]/[(n)(n-1)(n-2)/3]
Simplifying, we get:
(Tn+1)/(Tn) = (n+1)/(n-2)
Therefore, Tn+1/Tn = (n+1)/(n-2).