MathDoofus wrote:As promised, here's what I've got so far in using pen and paper and working back through the original problem in light of everyone's help.
I want to show the following: 1 + 2 + 3 + ... + n = n(n + 1)/2
First, I'll perform a check using 1 as a base case, to create the first "rung" on the ladder that I'm hoping will result.
1 = 1(1 + 1)/2
1 = 1(2)/2
1 = 2/2
1 = 1
The base case checks out.
Second, I'll assume that it's true for the k case. In other words, I'll use k as a placeholder for any natural number, and see what happens.
1 + 2 + 3 + ... + k = k(k + 1)/2
Again, the above is an assumption. I'm using it so that I can show that if something is true of the k case, then it's also true of the (k + 1) case.
Third, I'll add (k + 1) to both sides of the above-referenced equation. The reason that I can be absolutely sure that the (k + 1) version is equal to the version containing only k is because I'm adding an identical element (k + 1) to both sides of an equation.
1 + 2 + 3 + ... + k + (k + 1) = k(k + 1)/2 + (k + 1)
Do I have it all right so far?
Everything to this point is correct. Remember, k is just a placeholder for some number, so (k+1) is just a placeholder for some slightly different number. If you add the same number to both sides of an equality, it stays equal.
The reason you add (k+1) to both sides:
1 + 2 + 3 + ... + k = k(k + 1)/2 is an equation whose left-hand side is just the sum of the first k positive integers. We can turn it into the sum of the first (k+1) integers by adding the (k+1)th integer, which is just (k+1). But if we add (k+1) to the left-hand side, we have to add it to the right-hand side, too, to make sure both sides remain equal.
Why do we want to do this? Our goal is to say that, if we assume some fact about the k case is true (namely, that 1 + 2 + 3 + ... + k = k(k + 1)/2), then it's also true of the (k+1) case. We're interested in the sum of the first k integers; we are assuming 1 + 2 + 3 + ... + k = k(k + 1)/2. The (k+1) case is the sum of the first (k+1) integers, which we get to by adding (k+1) to both sides.