The magic of induction

academics induction mathematics

What is the sum of the first n positive integers? Phrased mathematically: 1 + 2 + 3 … + n -1 + n = ?. The answer, it turns out, is n * (n + 1) / 2. How do we show this is true though? How do we prove this?

One way is to dive head first into the sum and manipulate it so that we arrive at the answer. What can we do with: 1 + 2 +… + n -1 + n ? We can pair opposite sides of this sum and add those! 

1 - - - n = n + 1

2 - - - (n - 1) = n + 1

3 - - - (n – 2) = n + 1




n/2 - - - (n/2 + 1) = n + 1

Note, these all add to n + 1. And there are n/2 pairs! That gives us a total of n/2 * (n+1) = n(n+1)/2 !

At this point, some readers might be thinking: “Wait… what if n is odd? Then we won’t be able to pair all the numbers up.” Good point. If n is odd, we’ll have the middle number left over. But, the middle number will be (n + 1)/2. And, there will only be (n – 1)/2 pairs. So, the total remains the same.

(End Proof)

Whew, that was a little tedious and took some ingenuity.

Notice, with the above technique, we actually derived the answer. But, we were given the answer to start with! Is there a way to show a formula is true without deriving it? Cue, induction!


Induction is a technique for proving that a formula is true for all positive integers. It asks us to show that a formula has a particular property: if the formula is true for some number (call it k), it is also true for the next number, (k + 1). After this, all we need to do is show that the formula is true for k = 1. What follows is magic. Because the formula is true for k = 1, it is also true for k+1 = 2. Then, because it's true for 2, it's true for 3, and so on. Induction.

Let's try this out on our formula for the sum of the first n positive integers.

Step 1: Show that if the formula is true for some number k, it’s also true for the next number, k+1.


Imagine that for some number k, 1 + 2 + 3 … + k-1 + k = k(k + 1)/2. Imagine that this is true. Given this, does 1 + 2 … + k-1 + k + k+1 = (k + 1)(k + 2)/2 ? That is, is the formula true for k+1? 

What can we do with 1 + 2 … + k-1 + k + k+1 ? We can substitute in the formula for the sum of the first k numbers! So, 1 + 2 … + k-1 + k + k+1 = k(k + 1)/2 + k+1 = k(k + 1)/2 + 2(k + 1)/2 = (k+2)(k+1)/2. 


Step 2: Show that the formula is true for k = 1.


This’ll be short and sweet. What is the sum of the first 1 number? 1. Does this = 1 * (1 + 1)/2 = 1 * (2)/2 = 1. Yes.

To conclude, because the formula is true for k = 1, it’s true for k + 1 = 2. And because it’s true for 2, it’s true for 3, and so on.

(End proof)

Wow, I don’t know about you, but that felt like no work compared to the first proof. That’s what I call magic.

In conclusion

Induction is even more powerful when applied to more complicated formulas like the sum of the first n cubes, which require sophisticated techniques to derive.

I learned induction in high school and used it mechanically for years before I fully understood and appreciated it. Now, I honestly feel like it’s one of the most awesome things I’ve ever learned. I hope I conveyed some of my excitement for induction to you in this article. Keep proving.


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