]]>
LearnNext
Get a free home demo of LearnNext

Available for CBSE, ICSE and State Board syllabus.
Call our LearnNext Expert on 1800 419 1234 (tollfree)
OR submit details below for a call back

clear

Application of Mathematical Induction

5,665 Views
Have a doubt? Clear it now.
live_help Have a doubt, Ask our Expert Ask Now
format_list_bulleted Take this Lesson Test Start Test

Application of Mathematical Induction - Lesson Summary

Principle of Mathematical Induction

Let P(n) be a mathematical statement, where n is a natural number, such that:

(i) The statement is true for n = 1, or P(1) is true.

(ii) If the statement is true for n = k, where k is a positive integer, then the statement is also true for n = k + 1.


P(k) is true ⇒ P(k + 1) is true.

Then, P(n) is true for all natural numbers n.

Principle one is a statement of fact, while principle two is a condition.

If P(n) is true for all n ≥ 2, then step 1 starts from n = 2 and we verify the result of P(2).


Next, if the second principle is true for n = k, then it is also true for n = k + 1.

Using the principle of mathematical induction, prove the following statements:

(i) 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1)

(ii) For all n ∈ N; (23n - 1) is divisible by 7.


(i) 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1)

P(n): 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1)

For n = 1,

P(1): 1/1.3 = 1/((2x1)+1) ⇒ 1/3 = 1/3

∴ P(1) is true.

Assume that P(n) is true for some k ∈ N.

P(k): 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2k-1)(2k+1) = k/(2k + 1) ----------- (1)


To prove that P(k + 1) is true, add the next term of the series to both sides of the statement.

1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2k-1)(2k+1) + 1/([2(k+1)-1][2(k+1)+1])= k/(2k + 1) + 1/([2(k+1)-1][2(k+1)+1])

= k/(2k + 1)  +  1/((2k + 1)(2k + 3))

= (k/(2k + 3) + 1)/((2k+1)(2k+3))

=(2k2+3k+1)/((2k+1)(2k+3))

=((2k+1)(k+1))/((2k+1)(2k+3))

=(k+1)/(2k+3)

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

=P(k+1)

∴ P(k+1) is true.

Hence, P(n) is true for all natural numbers n.


We have to prove that the expression is divisible by seven for all natural numbers n.

P(n): (23n - 1) is divisible by 7.

For n = 1, P(1): 23x1 - 1 = 7, which is divisible by 7.

∴ P(1) is true.

Let P(n) be true for some k ∈ N.

∴ P(k):(23k -1) is divisible by 7 ------- (1)

Need to prove that P(k + 1) is true.

P(k + 1): 23(k+1) - 1

= 23k+3 - 1

=23k.23 - 1

=23k.8 - 1

=23k.(7+1) - 1

=23k.7 + (23k - 1)

23k.7is a multiple of 7.

23k-1 is divisible by 7 from equation 1.


∴ P(k+1) is divisible by 7.

Hence, (23n - 1) is divisible by 7 for all natural numbers n.

Comments(0)

Feel the LearnNext Experience on App

Download app, watch sample animated video lessons and get a free trial.

Desktop Download Now
Tablet
Mobile
Try LearnNext at home

Get a free home demo. Book an appointment now!

GET DEMO AT HOME