Create a proof by mathematical induction that demonstrates that the sum of the first n even numbers is equal to n(n+1).
Look daunting? No worries! At the heart of proof by induction is a simple truth. You know something is true for a particular math problem, and you see a pattern that could be true of all numbers. So, you show that the pattern holds true for next number, and the next, and so on, because you could never actually take the time to prove the entire number line.
Specifically, let me tell you what the formal proof below communicates. I am going to multiply 2 by 1, 2, and 3--then I am going to add those products (like so): 2(1) + 2(2) + 2(3). We know this is 2 + 4 + 6 = 12. The last number that I multiplied by 2 was the number 3. I see this pattern that I can then take the 3 and multiply it by the next number, which is 4. Turns out that 3 times 4 is also 12. For this whole pattern, 3 was the important number. Now I want to prove it goes on forever. That is why you see the "..." which continues the pattern and the 2 times n? The n just means you can keep on trying this pattern for as long as you want.
Forever and ever, you can multiply 2 by the counting numbers and add up all of those products. The largest number that you multiplied by 2 looks at the pattern on the other side. If the last number that you multiplied by 2 was a 4 (that was quick), then 2 + 4 + 6 + 8 = 4 (5) = 20.
In other words P(n) is true for 2+4+6+...+2n = n(n+1).
Solution: Let P(n) be the proposition that the sum of the first n even numbers is n(n+1). We must prove that P(n) is true for n = 1,2,3,....First, we need to show the basis step that P(1) is true. Then we show the inductive step where the conditional statement with a new variable P(k) implies P(k+1) is true for k = 1,2,3,.... That's the 2(1) + 2(2) + 2(3) = 3(4) = 12 with the conditional k in place of the proposition n.
Basis Step: P(1) is true, because 2(1) = 1(1+1). 2(1)=1(2) because of the commutative property of multiplication.
Also 2(1) + 2(2) + 2(3) = 2(1+2+3) = 2(6) = 2(2)(3) = 4(3) = 3(4). (This is distributive and commutative property, but I digress...)
Inductive Step: For the inductive hypothesis we must assume that P(k) holds true for any positive integer k. We assume that P(k) = 2+4+6+...+2k = k(k+1).
The forever and ever part...
Now we are going to start with that forever and ever part...and take it one step further. It is like the next stept of 2(1) + 2(2) + 2(3) + 2(4) = 4(5) = 20 statement. Basis Step - It must be shown that P(k+1) = 2+4+6+...+ 2k + 2(k+1) = (k+1)((k+1)+1) = (k+1)(k+2) is also true.
Induction Step - When we begin with the assumption P(k) = 2+4+6+...+2k = k(k+1) and add 2(k+1) to both sides, it follows that 2+4+6+...+2k + 2(k+1) = k(k+1) + 2(k+1) and when you factor out the (k+1) = (k+1)(k+2). Which, from the basis step, is P(k+1). [Factoring out a binomial can be tought. It is sort of like saying kx +2x = x(k+2) kind of factoring.]
This last equation shows that P(k+1) is true under the assumption that P(k) is true. This completes the inductive step.