1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, ...

Most of the proofs I've done with triangular numbers just involved some simple algebra. We just would perform the necessary operations to the explicit formula for triangular numbers n(n+1)/2 and it would simplify to the answer we wanted.

Today, I will be doing something very different that can't be proved with this method. Rather than looking at triangular numbers, let's look at cube numbers. Here are the first few:

1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, ...

Let's just add up the cubes. We will start by adding the first one, then the first two, the first three, and so on.

1 = 1

1 + 8 = 9

1 + 8 + 27 = 36

1 + 8 + 27 + 64 = 100

1 + 8 + 27 + 64 + 125 = 225

1 + 8 + 27 + 64 + 125 + 216 = 441

Do you notice any pattern with the sums? If you look closely, you'll see that they are all square numbers. But not any square numbers. Look again.

1 = 1 = (1)^2

1 + 8 = 9 = (3)^2

1 + 8 + 27 = 36 = (6)^2

1 + 8 + 27 + 64 = 100 = (10)^2

1 + 8 + 27 + 64 + 125 = 225 = (15)^2

1 + 8 + 27 + 64 + 125 + 216 = 441 = (21)^2

They are squares of the triangular numbers. When I first saw this, I was stunned that this happened, and thought there was no way this could continue to work forever. But in fact, we can prove it.

You could try doing it algebraically, but we can't represent cube numbers in a form that can be simplified. For this, we have to use something that I've mentioned before called proof by induction.

We already know it works for the first number. We know that:

1^3 = 1^2

or

1^3 = (1(1+1)/2)^2

Remember that triangular numbers can be found with the formula n(n+1)/2. Since we are squaring it, let's just simplify that.

(n(n + 1)/2)^2

((n(n + 1))^2)/((2)^2)

n^2(n + 1)^2/4

We know it works at some point, that is when n = 1, we can state our induction hypothesis. This is saying that for some number k, we know it will work.

1^3 + 2^3 + ... + (k - 1)^3 + k^3 = k^2(k + 1)^2/4

Now, we want to find out if we continue forward, it will keep working. In other words, we want to know if it holds true for k+1.

1^3 + 2^3 + ... + (k - 1)^3 + k^3 + (k + 1)^3 = (k + 1)^2(k + 2)^2/4

Based on our induction hypothesis, we know what the sum of the first k cubes is. So, we can substitute that into the new equation to get:

**1^3 + 2^3 + ... + (k - 1)^3 + k^3**+ (k + 1)^3 = (k + 1)^2(k + 2)^2/4

**k^2(k + 1)^2/4**+ (k + 1)^3 = (k + 1)^2(k + 2)^2/4

If you look on the left side, you will see that both terms have a factor of (k+1)^2. The distributive property says that we can "factor out" or restructure the equation so that the (k+1)^2 is multiplied at a different time. This gives us:

k^2(k + 1)^2/4 + (k + 1)^3 = (k + 1)^2(k + 2)^2/4

(k + 1)^2 • (k^2/4) + (k + 1)^2 • (k + 1) = (k + 1)^2(k + 2)^2/4

((k + 1)^2)(k^2/4 + k + 1) = (k + 1)^2(k + 2)^2/4

Now, rather than having a denominator on just one term, let's create a common denominator of 4.

((k + 1)^2)(k^2/4 + k + 1) = (k + 1)^2(k + 2)^2/4

(k + 1)^2(k^2 + 4k + 4)/4 = (k + 1)^2(k + 2)^2/4

The two sides are almost identical. The only thing that is different is the (k + 2)^2. But we can simplify that.

(k + 2)^2

(k + 2)(k + 2)

(k)(k) + (k)(2) + (2)(k) + (2)(2)

k^2 + 2k + 2k + 4

k^2 + 4k + 4

Let's substitute that in for (k + 2)^2.

(k + 1)^2(k^2 + 4k + 4)/4 = (k + 1)^2

**(k + 2)^2**/4

(k + 1)^2(k^2 + 4k + 4)/4 = (k + 1)^2

**(k^2 + 4k + 4)**/4

And now the two sides are equal. So, when all of the dust settled, it showed that the sum of the first k+1 cubes equalled the square of the k+1st triangular number, or the sum of the cubes is the square of the sum. In other words, the sequence continued to work from k to k+1 meaning that whatever value k is, the next value will hold true as well.

This proof is very complicated, with factoring, substituting and simplifying. But if you look at it a couple of times, it makes perfect sense and makes the fact even cooler.

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