Answer :

Given; P(n) = 7n – 2n is divisible by 5.

P(0) = 70 – 20 = 0; is divisible by 5.


P(1) = 71 – 21 = 5; is divisible by 5.


P(2) = 72 – 22 = 45; is divisible by 5.


P(3) = 73 – 23 = 335; is divisible by 5.


Let P(k) = 7k – 2k is divisible by 5; 7k – 2k = 5x.


P(k+1) = 7k+1 – 2k+1


= (5 + 2)7k – 2(2k)


= 5(7k) + 2 (7k – 2k)


= 5(7k) + 2 (5x); is divisible by 5.


P(k+1) is true when P(k) is true.


By Mathematical Induction P(n) = 7n – 2n is divisible by 5 is true for each natural number n.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Prove that cos α RD Sharma - Mathematics

Prove that sin x RD Sharma - Mathematics

Prove the followiRD Sharma - Mathematics

Prove the followiRD Sharma - Mathematics

Prove the followiRD Sharma - Mathematics

Prove the followiRD Sharma - Mathematics

Prove the followiRD Sharma - Mathematics

Prove the followiRD Sharma - Mathematics

Prove the followiRD Sharma - Mathematics

Prove that <span RD Sharma - Mathematics