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 PRINCIPLE OF MATHEMATICAL INDUCTION

  The following steps proof equation or sequence is a mathematical induction. 

Step 1; the true of a statement for ∀ n = 1

Step 2; we assume that the statement is true for n = k

Step 3; we prove the statement is true for n = k+1

Definition:

What is Mathematical Induction 

The three steps given serve as the key to show and prove MATHEMATICAL INDUCTION. Let's try some examples, so that we can clearly understand.

1 + 2 + 3 +.........+n = n(n+1)/2

 

Step 1; n = 1

1 = 1(1+1)/2 

1= 1(2)/2

1= 2/2

1 = 1 therefore n = 1

Step 2; n = k

1+2+.....+ k = k(k + 1)/2

Step 3; n = k+1 but this is containing both step 2 and 3 inside the inner sequence ,

In this we might directly find the solution without using the lcm method

Therefore we have

1 + 2 +........+k(k +1) = k + 1 ( k + 1 + 1)/2

 = [k (k + 1) + k + 1 + k + 1]/2

= [k (k + 1) + 2(k + 1)]/2

Expand the brackets

= [ k² + k + 2k + 2]/2

 by Collecting the like term we have.

= [k² + 3k + 2 ]/2

By applying factorization we have;

= [(k + 1) (k + 2)]/2

= [ k +1 (k + 1 + 1)]/2

Therefore , but  n = k+1 

n (n + 1)/2.     Hence proved.

Example 2; 

Show that all positive integer ℤ⁺ n is a mathematical induction.

1 + 2 + 3 +..........+ n² = [n ( n +1)(2n + 1)]/6

Step 1; n = 1 

=[ 1(1+1)(2(1)+1)]/6

=[1(2)(3)]/6

=2*3 = 6

Therefore 6/6 = 1

Step 2; n = k

[k ( k + 1)(2k + 1)]/6

Step 3; n = k + 1

k (k+1)² = [k (k+1)(2k+1)]/6+ (k+1)²/1

By taking the Lowest common multiple lcm

= [k (k+1)(2k+1) + 6(k+1)(k+1)]/6

They are likely event and so we cancel them.

= [k (2k+1) + 6(k+1)]/6

We can now take the multiplication.

=[ 2k² + k + 6k + 6]/6

Simplify Using factorization

=[ (k + 1)(k + 2)(2k + 3)]/6

Change it to mathematical Induction format.

=[ (k+1)(k+1+1) 2(k+1)+1]/6

=[ n(n+1)2n+1]/6

Example 3;

 Prove the Geometric Progression sequence using mathematical Induction.

a + ar + ar² + ...... + arⁿ⁻¹ = a(1 - rⁿ)/1 - r

Show that the above sequence and equation are mathematical Induction

To show we Introduce some various steps

Step 1; n = 1

ar¹⁻¹ = a (1 - r¹)/1 - r 

Using the law of Indices which says whenever division is between the powers should be subtracted.

After subtracting you will learn that the powers are turning into zero. And another law of Indices says anything rest to power zero is equal to one .

Therefore the entire solution is equal to a(1)= a

Step 2; n = x

arˣ⁻¹ = a (1- rˣ)/ 1- r

Step 3; n = x + 1

a (1 - rˣ)/1 - r + arˣ⁺¹⁻¹/1

By applying lcm we can have 

[a(1 - rˣ) + arˣ](r)/(1 - r)

Expand the brackets

[ a - arˣ + arˣ](r)/(1 - r )

Let's rewrite.

[a - arˣ - arˣ + arˣ](r)/(1 - r)

a -  arˣ(r)/(1 - r)

a(1 - rˣ⁺¹)/(1 - r)      hence proved

  EXERCISES ON MATHEMATICAL INDUCTION

 1 × 3 + 2 × 4 +.......+ n(n+2) = [n(n+1)(2n+7)]/6

  3 + 8 +....+ n² - 1 =[ n(n-1)(2n+5)]/6

  1/1×2 + 1/2×3 + 1/n(n+2) = n/n+1

HISTORY OF MATHEMATICAL INDUCTION 

The first known use of mathematical Induction is within the work of the sixteenth century mathematian name Francisco maurolico (1494 - 1575).

Francisco wrote extensively on the works of classical mathematics and also gave a great contribution to geometry and optics.

 During this boom arithmeicorum, lib-duo, Francisco present a spread of properties of the integers with proof of those properties.

To prove some of this properties he deviced the use of mathematical Induction .

His first use of mathematical Induction during this book was to prove that the sum of the primary n odd positive integers equals n².

Augustus De Morgan is credited with the primary presentation at 1838 formal proof using mathematical Induction, also as introducing the terminology."mathematical Induction". Francisco mathematical proof was informal and he never use the word " Induction". The name  Induction was employed by an English Mathematician John Wallis

Remember how mathematical Induction works, imagine the infinite ladder and therefore the rules of reaching steps can assist you understand and visualise how mathematical Induction works.

Important notice: statement [1] and [2] for the infinite ladder are precise the basis steps and inductive step, respectively.

The prove that P(n) is true for all positive integer n, where P(n) is that the statement that will reach the nth rung of the ladder.

Why is Mathematical Induction a legitimate proof technique.

Why Mathematician tons use mathematical Induction for proving result. The rationale comes from well ordering property of the induction. As an example, the set of positive integers, which state that each nonempty subset of the set of positive integers features a lest element. Therefore suppose we all know that P(1) is true which the preposition P(k) ⇒ P(k+1) is true for all positive integer k.

Many theorem have been proved using mathematical Induction, like equation or formulae often interpreted in geometrically, algebraic stand point, probabilistically, and machine learning optimisation stand point. That including mathematical Induction.

 

CONCLUSION ON MATHEMATICAL INDUCTION

Key point of mathematical Induction is reduction, generalisation depends on working with different kind of cases and developing a conjecture by observing incidences till we have observed each and each case.

Therefore in simple and straightforward we can say that "Induction" means generalisation.

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