Principle of Mathematical Induction

Principle of Mathematical Induction:-

<img src="principle of mathematical induction" alt="principle of mathematical induction">

we know that the first positive even integer is 2 = 2  \times 1
the second positive even integer is 4 =  {\rm{ }}2 \times 2

the third positive even integer is 6 = 2 \times 3
the fourth positive even integer is 8 = 2 \times 4

………………………………………………….

If we move using the same pattern then nth positive integer=2n

but this is just the summary of the above observations and it’s a statement which we believe is true. We will call these types of statements “proposition”. A proposition will remain a proposition until we prove it true. We represent a proposition by a symbol {{\rm{P}}_{\rm{n}}} . 

PRINCIPLE OF MATHEMATICAL INDUCTION(PMI)- “Principle of Mathematical induction proves that we can climb on a ladder as high as we wish, by proving that we can climb the first rung/bottom rung (the base) and from each rung, we can go up to the next rung”

If we assume a mathematical proposition({{\rm{P}}_{\rm{n}}}) to be like a ladder we can say that

(i) If {{\rm{P}}_{\rm{n}}}  is true for the first term where  n \ge Z (this is called base step)

 

Also READ this-  Increasing and Decreasing Functions

(ii) If {{\rm{P}}_{\rm{n}}}  is true for the (k+1) terms while it’s already true for k terms where  n \ge Z (this is called induction step)

Then according to Principle of Mathematical Induction,{{\rm{P}}_{\rm{n}}}  will be true for all  n \ge Z

Practical Example of PMI (Domino Effect) -:

<img src="domino effect" alt="domino effect">Domino effect is one of the best examples of PMI

(i) If the first domino hits the second one then second will hit the next

(ii) If k-th dominos falls the (k+1)th will also fall

This approves the Principle of Mathematical Induction

Numerical Example of (PMI) :-

Prove that 1 2 + 2 2 + 3 2 + … + n 2 =  \frac{{n\left( {n{\rm{ }} + {\rm{ }}1} \right){\rm{ }}\left( {2n{\rm{ }} + {\rm{ }}1} \right)}}{6}  is true where  n \in N

Solution:- To prove this statement we use the principle of mathematical induction

  • Base Step: We first check if p (1) is true. L.H.S = 1 2 = 1R.H.S = 1 (1 + 1) (2 \times 1 + 1)/ 6  = 1
  • Both sides are equal so for n=1 it is true. 
  • Induction Step: We now assume that p (k) is true 1 2 + 2 2 + 3 2 + … + k 2 = k (k + 1) (2k + 1)/ 6 ……(i)
  • now we will check for p (k + 1). So we  add (k + 1)th term to both sides of the above statement 1 2 + 2 2 + 3 2 + … + k 2 + (k + 1) 2 = k (k + 1) (2k + 1)/ 6 + (k + 1) = (k + 1) [ k (2k + 1)+ 6 (k + 1) ] /6
  • now we simplify this  = (k + 1) [ 2k 2 + 7k + 6 ] /6= (k + 1) [ (k + 2) (2k + 3) ] /6
  • We have begun with P(k) and we have proved that1 2 + 2 2 + 3 2 + … + k 2 + (k + 1) 2 = (k + 1) [ (k + 2) (2k + 3) ] /6
  • Which is the statement P(k + 1).

Ib Elite Tutors provide the best Ib online tutors for all subjects. Fill this form and secure your free demo session

Also READ this-  IB Tutors-useful pdfs

 

<img src="demo" alt="demo">