Prove that "6^n + 9" is divisible by 5 for all natural numbers.

First assess that the initial case of where n = 1 is true. In this case, 6+9=15=53, so we can see that the case is true.We can then assume that 6k+9 is a multiple of 5, so we can let 6k+9 = 5A for some A in the natural numbers. We then consider the case of n = k+1, so consider 6k+1+96k+1+9 = 66k+9 = (6k+9) + (5*6k) = 5(A+6k) So it must be a multiple of 5The problem is shown true for the case of n = 1, and by assuming it is true for some k, it is shown to be true for the case n = k+1. So by the principle of mathematical induction it is true for all natural numbers n.

Answered by Further Mathematics tutor

2821 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Find the complementary function to the second order differential equation d^2y/dx^2 - 5dy/dx + 6x = x^2


Why am I learning about matrices? What are they?!


Write 1 + √3i in modulus-argument form


Understanding differentiation from first principle.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences