MathDoofus wrote:But how does "assume A, therefore B" prove anything?

It proves that *if* you are on some rung of the ladder, *then* you can step up to the next rung.

You are right that it doesn’t say anything about whether you can reach the rung you are assuming to start on, it just says that *if* you start there, then you can get to the next rung.

That is *all* you are trying to prove with the induction step. That *if* your statement is true about k, *then* it is also true about (k+1).

Suppose that you are able to do so. What does that mean? Well it means you can plug in any k and say “If my statement is true about k, then it is also true about (k+1).”

Where does that get you? Well, you can plug in n=4 and say, “If my statement is true about 4, then it is also true about 5.”

And, “If my statement is true about 5, then it is also true about 6.”

And, “If my statement is true about 6, then it is also true about 7.”

And so forth.

MathDoofus wrote:Here's where I'm confused:

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

We take this as a given. We assume it is true for some k.

MathDoofus wrote:Here's where I mash up k and (k + 1), for some reason, which strikes me as assuming both k and (k + 1).

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

All you have done here is add (k+1) to both sides of the previous equation. Perfectly valid.

MathDoofus wrote: Here's another part that feels like cheating, where I substitute the right side of the first equation for part of the left side of a later one.

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

Below is where I get stuck on the algebra and how to "solve" this equation.

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

Ah, you seem to misunderstand the goal here. You are not trying to “solve” that equation. You are trying to show that the thing you started with, (1+2+3+…+k+(k+1)), has the same value as the thing you *think* it should equal, namely your statement with (k+1) plugged in.

In other words, you need to manipulate (1+2+3+…+k+(k+1)) with valid operations which do not change its value, until it ends up in the form n(n+1)/2 with n=(k+1).

And the only thing you know about k is that (1+2+3+…+k) = k(k+1)/2.

You are trying to show that *if* the formula works for n=k, *then* the formula also works with n=(k+1). That is how you step up the ladder. Assume you are on rung k, and prove that *if* you are on rung k, *then* you can reach rung (k+1).