"I think it would be worthwhile to explicitly ca..."

We'll do an example to build our intuition before giving the proper definition of the principle\. We'll provide yet another proof that $$1 + 2 + \\cdots + n \= \\frac{n(n+1)}{2}$$ for all integers $n \\ge 1$\. First, let's check the base case, where $n\=1$: $$1 \= \\frac{1(1+1)}{2} \= \\frac{2}{2} \= 1.$$ This is \(fairly obviously\) true, so we can move forward with the inductive step\. The inductive step includes an assumption, namely that the statement is true for some integer $k$ that is larger than the base integer\. For our example, if $k\\ge1$, we assume that $$1 + 2 + \\cdots + k \= \\frac{k(k+1)}{2}$$ and try to prove that $$1 + 2 + \\cdots + k + (k+1) \= \\frac{(k+1)(\[k+1\]+1)}{2}.$$ Take our assumption and add $k+1$ to both sides: $$1+2+\\cdots + k + (k+1) \= \\frac{k(k+1)}{2} + k + 1.$$ Now the left\-hand sides of what we know and what we want are the same\. Hopefully the right\-hand side will shake out to be the same\. Get common denominators so that the right\-hand side of the above equation is $$\\frac{k(k+1)}{2} + \\frac{2(k+1)}{2} \= \\frac{(k+2)(k+1)}{2} \= \\frac{(k+1)(\[k+1\]+1)}{2}.$$ Therefore, $$1 + 2 + \\cdots + k + (k+1) \= \\frac{(k+1)(\[k+1\]+1)}{2}$$ as desired\.

I think it would be worthwhile to explicitly call out that what's happening here is that we're replacing $n$ in the original equation with $k+1$.