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.

Related Further Mathematics A Level answers

All answers ▸

Why does e^ix = cos(x) + isin(x)


Find the four roots of the equation z^4 = + 8(sqrt(3) + i), in the form z = r*e^(i*theta). Draw the roots on an argand diagram.


Integrate (4x+3)^1/2 with respect to x.


Find roots 'a' and 'b' of the quadratic equation 2(x^2) + 6x + 7 = 0


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