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$~$.