MathDoofus wrote:WibblyWobbly wrote:For this proof to work, the above equation needs to be equivalent to the (k+1)-case:
1 + 2 + 3 + ... + k + (k + 1) = [(k+1)((k+1)+1]/2
Well, we can see that the left hand sides of those two equations are equal. The way we prove that
1 + 2 + 3 + ... + k + (k + 1) = k(k + 1)/2 + (k + 1) is equivalent to 1 + 2 + 3 + ... + k + (k + 1) = [(k+1)((k+1)+1]/2
is by showing that the right hand sides are the same:
k(k + 1)/2 + (k + 1) = [(k+1)((k+1)+1]/2
Yes, that may be the source of the confusion - so why is it that, when I plugged in (k + 1) to both sides, I wasn't actually writing the (k + 1) case? And what relationship does substituting (k + 1) for every instance of k have to (1) the k case (which is Step 1/Equation 1) and (2) the case where k and (k + 1) are related (which is Step 2/Equation 2)?
First, some terminology. You didn't "plug in" (k+1) to both sides, you added it to both sides. "Plugging in" is the same as substitution, and you don't want to substitute anything here. Now, when you added (k+1) to both sides, what you're doing is actually writing the (k+1)-case directly from the k-case - it just looks a little different than if you had substituted (k+1) for n in the original equation (which is also the (k+1)-case). That's actually what we're trying to prove: that adding (k+1) to both sides of the k-case equation gives you the equation for the (k+1)-case.