32 2. Background The format of the proof above is typical of inductive proofs of summation formulas: use the inductive hypothesis to simplify a por- tion of the next value’s sum. For the next example we need to know a convex polygon is one where all the corners point out. If you connect two corners of a convex polygon with a straight line segment, the segment will lie entirely within the polygon, cutting it into two smaller convex polygons. As you get more comfortable with induction, you can write it in a more natural way, without segmenting off the base case and inductive step portions of the argument. We’ll do that here. Notice the base case is not 0 or 1 for this proof. Example 2.5.3. For n 2, the sum of angle measures of the interior angles of a convex polygon of n vertices is (n 2) · 180◦. Proof. We work by induction. For n = 3, the polygon in question is a triangle, and it has interior angles which sum to 180◦ = (3−2)·180◦. Assume the theorem holds for some n 3 and consider a convex polygon with n + 1 vertices. Let one of the vertices be named x, and pick a vertex y such that along the perimeter from x in one direction there is a single vertex between x and y, and in the opposite direction, (n +1) 3 = n 2 vertices. Join x and y by a new edge, dividing the original polygon into two polygons. The new polygons’ interior angles together sum to the sum of the original polygon’s interior angles. One of the new polygons has 3 vertices and the other has n vertices (x, y, and the n 2 vertices between them). The triangle has interior angle sum 180◦, and by the inductive hypothesis the n-gon has interior angle sum (n 2) · 180◦. The n + 1-gon therefore has interior angle sum 180◦ + (n 2)180◦ = (n + 1 2) · 180◦, as desired. Notice also in this example that we used the base case as part of the inductive step, since one of the two polygons was a triangle. This is not uncommon. Exercise 2.5.4. Prove the following statements by induction. (i) For every positive integer n, 1 + 4 + 7 + . . . + (3n 2) = 1 2 n(3n 1).
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