In Week 9, looked at three questions concerning the sum of cubes and the square of sums. (If you haven’t had a chance to look at them yet, click here.) The three questions build upon each other — and in the end, we get a little proof of one of number theory’s famous results.

1. We know that 1 + 2 + 3 + …. + k = k(k+1)/2. We can add k + 1 to both sides of the equation. Then we have

1 + 2 + 3 + …. + k + (k+1) = k(k+1)/2 + (k+1)

Now we can combine the terms on the righthand side.

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

Therefore, 1 + 2 + 3 + …. + k + (k+1) = (k+1)(k+2)/2

2. We’re given that 1^3 + 2^3 +….+k^3 = k^2(k+1)^2/4, and we want to show that 1^3 + 2^3 +…. + k^3 +(k+1)^3 = (k+1)^2(k+2)^2/4.

We can add (k+1)^3 to both sides of the original equation without violating the equality.

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

As in the first problem, we can combine the terms on the righthand side, and then simplify the form.

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

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

Therefore, 1^3 + 2^3 +…. + k^3 +(k+1)^3 = (k+1)^2(k+2)^2/4

3. Now we can use the results from problems 1 and 2 to solve problem 3.

We want to show that (1 +2 + 3 + … + n)^2 = 1^3 + 2^3 + 3^3 + …. n^3.

In the first problem, we were able to show that

1 + 2 + 3 + …. + k + (k+1) = (k+1)(k+2)/2

In the second, we showed that 1^3 + 2^3 +…. + k^3 +(k+1)^3 = (k+1)^2(k+2)^2/4

What do you notice about these two statements?

(k+1)^2(k+2)^2/4, from the second question, is equivalent to squaring (k+1)(k+2)/2, from the first.

That means that (k+1)^2(k+2)^2/4 is also equivalent to squaring the lefthand side of the first equation.

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

Then we can substitute 1^3 + 2^3 +…. + k^3 +(k+1)^3 for (k+1)^2(k+2)^2/4, because we’ve already shown that they’re equivalent.

(1 + 2 + 3 + …. + k + (k+1))^2 = 1^3 + 2^3 +…. + k^3 +(k+1)^3. Our proof is complete.

This is what is known as a proof by induction.