💯 Mathematical Induction Divisibility Proof | 4^2n - 1 is Divisible by 5

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Timeline
0:00 Intro
0:44 Step 1 Checking for the initial value
1:19 Step 2 Assumption
1:46 Step 3 Proof
3:51 Conclusion

Base case: For n = 1, we have 4^(2 * 1) - 1 = 16 - 1 = 15, which is divisible by 5.

Inductive step: Assume that the result holds for n = k, i.e. 4^(2k) - 1 is divisible by 5. That is 4^(2k) - 1 = 5P, where P is any positive integer.
We need to prove that 4^(2(k+1)) - 1 is divisible by 5.
4^(2(k+1)) - 1 = 4^(2k) * 4^2 - 1 = (5P +, 1) * 4^2 - 1 = (5P +) 1 * 16 - 1 = (5P + 1) * 16 - 1 = 80P + 15 = 5(16p + 3), which is divisible by 5 where P is a positive integer (by the induction assumption). Therefore, 4^(2(k+1)) - 1 is divisible by 5.

Since the base case and inductive step hold, by mathematical induction, we can conclude that 4^(2n) - 1 is divisible by 5 for all positive integers n.