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 ▸

Find the integral of f(x)= x^3 + 2x^2 + 1


Given that p≥ -1 , prove by induction that, for all integers n≥1 , (1+p)^k ≥ 1+k*p.


By forming and solving a suitable quadratic equation, find the solutions of the equation: 3cos(2A)-5cos(A)+2=0


Let f(x)=x^x for x>0, then find f'(x) for all x>0.


We're here to help

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