The Domino Effect
Mastering Induction
Welcome, future engineers! Today, we are going to demystify one of the most elegant tools in your mathematical arsenal: Mathematical Induction.
Imagine a long line of dominoes stretching to infinity. To prove that every single one will fall, you only need to do two things: push the first one over, and ensure that if any domino falls, it knocks over the next one. That is exactly what we are doing with this divisibility problem.
Phase 1
The Base Case
We are tasked with proving that P(n):(25)n+1−24n+5735 is divisible by 242, which is 576, for all n∈N.
Our first step is the base case,
n=1. We substitute
n=1 into our expression:
P(1)=251+1−24(1)+5735
This simplifies to 252−24+5735, which is 625−24+5735=6336.
Now, we check if
6336 is divisible by
576. A quick division,
5766336=11
confirms it! The first domino has fallen. The statement holds for
n=1.
Phase 2
The Inductive Hypothesis
Now, we assume the statement is true for some arbitrary natural number k. This is our 'Inductive Hypothesis.'
We assume:
P(k)=25k+1−24k+5735=576m
for some integer
m.
Here is the pro-tip: do not just stare at this equation. Manipulate it! We want to isolate the highest power term, 25k+1, because it is the most difficult part of the expression.
By rearranging, we get:
25k+1=576m+24k−5735
Keep this in your back pocket; it is the key that will unlock the next step.
Phase 3
The Inductive Step
Now, we must prove that if
P(k) is true, then
P(k+1) must also be true. Let's write out
P(k+1) by replacing
n with
k+1:
P(k+1)=25(k+1)+1−24(k+1)+5735
This simplifies to
25k+2−24k−24+5735. To connect this to our hypothesis, we split the exponential term:
P(k+1)=25⋅25k+1−24k+5711
Now, we substitute our isolated expression from the hypothesis:
P(k+1)=25(576m+24k−5735)−24k+5711
Phase 4
The Victory Lap
Now, we expand and simplify. Distributing the
25, we get:
25⋅576m+25⋅24k−25⋅5735−24k+5711
This looks intimidating, but watch the magic happen. We have:
25⋅576m+600k−143375−24k+5711
Combining the k terms, 600k−24k=576k. Combining the constants, −143375+5711=−137664.
Our expression is now:
25⋅576m+576k−137664
We can factor out 576 from the first two terms easily. For the constant, 137664÷576=239.
Thus:
P(k+1)=576(25m+k−239)
Since m and k are integers, the term in the bracket is an integer. We have proven that P(k+1) is a multiple of 576.
By the Principle of Mathematical Induction, the statement is true for all n∈N. You have conquered the problem!