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

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.

Done!

### 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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