Question

Extended Principle of Mathematical Induction The Extended Principle of Mathematical Induction states that if Conditions I and II hold, that is,

(I) A statement is true for a natural number j

(II) If the statement is true for some natural number $k\ge j$, then it is also true for the next natural number $k+1$.then the statement is true for all natural numbers $\ge j$. Use the Extended Principle of Mathematical Induction to show that the number of diagonals in a convex polygon of n sides is $\frac{1}{2}n(n-3)$.

Short answer

The statement is shown.

Step-by-step solution

Step 1. Given information

Extended Principle of Mathematical Induction is given.

Step 2. Prove that the number of diagonals in a convex polygon of n sides is 12n(n-3).

First we consider the case of a square that has 4 sides and 2 diagonals.

So taking the value of n = 4 in above statement $\frac{1}{2}·4.(4-3)=2$, clearly shows that above statement is true for n = 4.

Hence condition 1 of Extended Principle of Mathematical Induction is true for n = 4

Now let us assume the given statement is true for some $k\ge 4$

Let us consider the case for k + 1 sides which can be made by placing a triangle on a side AB of polygon as shown. It has all diagonals of k sided polygon plus diagonals drawn from vertex M to all vertices of previous k sided polygon except for 2 , namely localid="1647848950539" $\mathrm{A}$
and B. In addition former side AB also becomes a diagonal.

Thus k + 1 sided polygon has a total number of diagonals as

$\begin{array}{rcl}\frac{1}{2}k(k-3)+(k-2)+1& =& \frac{k^2-3k+2(k-2)+2.1}{2}\\ & =& \frac{k^2-k-2}{2}\\ & =& \frac{(k+1)(k-2)}{2}\end{array}$

As we result we say that the given statement is true for all natural numbers n.