Prove by mathematical induction that, for all non-negative integers n, 11^(2n) + 25^n + 22 is divisible by 24

The procedure of mathematical induction is as following:

Firstly prove the base case is true i.e. when n=1, the statement is true.

Then assume for some integer n=k the statement is true, and then prove the case n=k+1, the statement is true.

Make a conclusion that by mathematical induction, the statement is true.

For this particular question, the base case is when n=0, the statement is true, since it is asked for 'all non-negative integers'. It is because 112x0+250+22=24 is divisible by 24 (24=24x1).

Then let's say P(n) is the proposition that 112n + 25n + 22 is divisible by 24. Assuming that P(k) is true for some integer k=n, then 112k + 25k+ 22  is divisible by 24.

The most important step comes: we then prove that P(k+1) is true. i.e. 112k+2 + 25k+1 + 22 is divisible by 24.

It is true because 112k+2 + 25k+1 + 22=121x 112k+25x25k+22=(120 + 1)112k + (24 + 1)25k + 22= (120 x112k + 24 x25k​)+ (112k + 25k+ 22). Expressions in both brackets are divisible by 24. so P(k+1) is true.

Then we are done. We could conclude that by mathematical induction the statement is true for all non-negative integers. 

Related Further Mathematics A Level answers

All answers ▸

A complex number z has argument θ and modulus 1. Show that (z^n)-(z^-n)=2iSin(nθ).


Prove by induction that 1^2 + 2^2 + 3^2 + . . . + n^2 = (1/6)n(n+1)(2n+1)


Find the values of x where x+3>2/(x-4), what about x+3>2/mod(x-4)?


Using graphs, show how the Taylor expansion can be used to approximate a trigonometric function.


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 2013–2024

Terms & Conditions|Privacy Policy