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

3111 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

What is sin(x)/x for x =0?


Prove by induction that (n^3)-n is divisible by 3 for all integers n>0 (typical fp1 problem)


I'm struggling with an FP2 First-Order Differential Equations Question (Edexcel June 2009 Q3) and the topic in general!


Integral of ln x


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning