Using Mathematical Induction to Prove 2^(3n) – 3^n is divisible by 5

Hello and welcome! In this video, we can use mathematical induction to prove that 2^(3n) – 3^n is divisible by 5 for all natural numbers, or n is greater than
or equal to 1 the first step in mathematical induction is
always the basis test which is to see if this condition or expression holds true for n=1 We put a question mark because we are just going to test if it is true or not, so… if n=1… then I get 2^(3*1) – 3^1… that’s equal to… 2^3=8 and 3^1=3 so I get 8 – 3 which is equal to 5. And
is 5 divisible by 5? Yes it is, so the basis test validates this condition. Now because we
have found that this statement is true for n=1, I’m going to assume that it holds true when n is equal to any integer so I’m going to assume true for n=k so I get 2^(3k) – 3^k=5J So this simply means that on the the
right hand side we have 5 multiplied by any integer or the right hand side basically means I get any multiple five, which is divisible by 5. So I say: J is any natural number. I know that this equation
can imply that 2^(3k)=5J + 3^k and this step will become important later on Ok now, the second major step of
mathematical induction is the inductive test. So I’ve assumed that if this statement is true
for n=k… does is hold true for when n=k + 1? or when I go to the next number? So if I substitute in n=k + 1… I’ll get 2^[3(k+1)] – 3^(k+1) and that can be written as 2^(3k+3) So I’ve just expand… distribute this out…
and I get 3^(k+1) and if I just use my index laws to simplify, I will get… (2^3k)(2^3) – (3k)(3) which is equal to 8(2^3k) – 3(3^k) alright, here’s where… this expression becomes important we have 8[5J + 3^k] – 3(3^k) If I expand that out, I will get 40J + 8(3^k) – 3(3^k) an well 8(3^k) – 3(3^k) that will give me 5(3^k)… and 40J remains… now if I factor out a 5, … I will get inside of brackets… 8J + 3^k well this term inside of the parentheses, or the brackets will always work out to be a whole number if J and k are natural numbers which they are and we have a whole number multiplied by 5 which means it is going to be divisible five. So we have proven by mathematical induction that the expression… 2^(3n) – 3^n is divisible by 5 for all natural numbers. So for all n is greater
than 1 and that’s all we’re required to do. So please give me a thumbs up if this video has helped you to better understand how to use mathematical induction. If you are math student please feel free to subscribe for future
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