Prove by mathematical induction that 11^n-6 is divisible by 5 for all natural numbers n

First I would do the base case (the first value):Test n=1,111-6=5. 5 is divisible by 5 therefore true for n=1.
Now we assume true for n=k,11k-6 is divisible by 5.Next we test n=k+1,11k+1-6We can rearrange this into 1111k-6= 1011k+11k-6We know that for n=k the result is 11k-6 which we assume to be true so that part can be assumed to be true.The first part can be factorised into 5(2*11k) which is divisible by 5. Therefore we have shown that if true for n=k, true for n=k+1 and as we shown true for n=1 it must also be true for all natural numbers. So we have proved this through induction

Answered by Further Mathematics tutor

3123 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

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


Using de Moivre's theorem demonstrate that "sin6x+sin2x(16(sinx)^4-16(sinx)^2+3)"


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.


3 points lie in a plane; P1=i+2j+3k, P2=-3i+5j+2k, P3=i+2j+k. Find the Cartesian equation of the plane


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