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Proof of Mathematical Induction


In simple terms Proof of Mathematical Induction is a process typically which is used for proving of the Math statement which is true for all positive integers. Perhaps we must not get confused with the proof by mathematical induction with the non-rigorous reasoning of mathematics, whereas it is a rigorous deductive reasoning of mathematics. We can understand this by taking an example;
0 + 1 + 2 + 3 + 4 + ····· + n = n (n + 1)/2. (This is true for all natural number).

Now we have to prove that this statement holds true for all the positive integers.
Put n = 1,
1 = 1(1 + 1) / 2,
As we can observe that on the left hand side of this equation there exists only single term that is 1 while on the right hand side if we are going to solve that part then the outcome will be 1. Now it is clear that both sides i.e. the left hand side (LHS) and the right hand side (RHS) are equal. Therefore this statement holds true to all n Є 1 (natural number belongs to 1) then it is clear that this statement will also be true for every n Є n+1.
Here two things should be noted that while conducting the mathematical induction proofs the proving takes place with two important steps, first one is the basic step which says that the statement must be true for all positive integers and then the other step says that which is a inductive step is, this statement will also hold true for all positive integers which belong to n + 1. This method of proving in math induction is helpful for proving the statement true and after the completion of first step the next step is automatically concluded, this may remind you of domino effect (this effect is a chain sequence effect which takes place with a small simple change), if the first one fails the other simply by itself gets discharged.

Sum of Odd Numbers

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Number system Numbers can be divided into two categories i.e. even numbers and Odd Numbers. Even numbers are numbers which when divided by 2 give zero as remainder and when we divide odd numbers by two we get remainder as 1. Now let us see what sum of odd numbers is:

1: First property of odd numbers: When we add two odd numbers then sum is always equals to an even number. For example:
1+ 3 = 4,
4 is an even number.
When we add even number of odd numbers we get resulting sum as an even number. For example:
3 + 5+ 7+ 9 = 24

2: When we add odd number of odd numbers then we get resulting sum as an odd number. For example:
1 + 3 + 5 = 9
9 is also an odd number.

3: There is a special property of sum of odd numbers that when we add first 'n' odd numbers then result is always a perfect Square number. For example:
1 + 3 = 4,
1 + 3 + 5 = 9,

4 and 9 both are perfect square numbers.
4: If we add first 'n' odd numbers which has the difference of 2 then we can calculate the sum of odd numbers by formula n2.
Sum of 'n' odd numbers = n 2.
If difference d = 2.

5: We can represent the sum of 'n' odd numbers by equation:
1 + 3 + 5 + 7 + . . . . . . + 2n – 1 = n 2.
This equation is a geometric interpretation of sum of odd numbers.

Counting Regions

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Counting is a crucial part of mathematics. There are several methods like Permutation, combination and Factorization which we follow to count the number of possibilities. Counting regions is also one such kind of problem we may find in Math.

Let us derive a formula to know maximum number of regions that are possible to be made by 'N' straight lines residing in a plane. Suppose we have 'N' lines in a plane with no line parallel to each other and so there are possibilities of their intersections. The regions that are formed by the Intersecting Lines can be evaluated as:

Based on our counting we find these number of regions being formed by each line as shown in the above figure:
Line 1: 2 regions
Line 2: 4 regions
Line 3: 7 regions
Line 4: 11 regions
Arranging these region counts in a particular order we see that they follow a particular fashion. Our sequence is 2, 4, 7, 11. By this we can easily derive the general formula for counting the regions for any number of lines we consider. The formula is:
Σ (N + 1) = (N (N + 1) / 2) + 1,
We have many applications of this concept in mathematics especially in geometrical problems. Let us see an example of it.

Example: In the following figure find out the total number of squares:

Solution: In figure we can see that there are 4 sides possible. Calculating number of squares for each side we get:
Number of squares that can be seen of side 1 is equals to 4 * 6 = 24. Likewise we have 15 squares of side 2, 8 of side 3 and 3 of side 4. Summing up all of them we get 50 squares in total in the given figure.