In normal induction we proved that if base case is true then we assume some number n to be true then we prove n+1 is true.
As written, this is inaccurate, though that may be simply a result of poor phrasing.
In standard induction, we do two things:
- First we prove the "base case". That the result holds for $n=1$.
- Then we prove that for every positive integer $n$, if the result holds for $n$, then it also holds for $n+1$.
(This is different from what you wrote; what you wrote is that one proves that if the result holds for $1$, then if it holds for some $n$, then it holds for $n+1$.)
In strong induction, we only need to prove one thing:
- For every positive integer $n$, if the result holds for all positive integers $k\lt n$, then the result holds for $n$.
This is enough to establish the result holds for all positive integer: if the result does not hold for all positive integers, then it fails to hold for some. Take the smallest integer $n$ for which the result does not hold. Then it holds for all strictly smaller integers; but by the implication above, this would imply that it holds for $n$ as well, a contradiction. So it holds for all positive integers.
(I used what is called the "Well Ordering Principle for Positive Integers": every nonempty collection of positive integers has a smallest element; this is in fact equivalent to induction).
Caveat: In strong induction, it is often the case that the general argument proving the implication does not hold for all $n$, but only for all "sufficiently large" $n$. In that case, we need to establish the implication for some $n$ "by hand". This is often, incorrectly, called a "base" of the induction. In fact, it is a "special case" of the proof of the single inductive step.
Here, the proposition being proven is:
Either $n\lt 12$, or else an $n$ cent stamp can be made using only $3$- and $7$-cent stamps.
The strong inductive step is:
Assume that the result is true for all $k$ strictly smaller than $n$. Then it holds for $n-3$; we can make an $n-3$-cent stamp using $3$- and $7$-cent stamps. Then we can make an $n$-cent stamp by making an $n-3$-cent stamp, and adding a $3$-cent stamp. So we can make an $n$-cent stamp. QED
The problem is that this argument works if $n$ is "sufficiently large", but it does not work if $n\lt 12$ (because that is not what we need to prove for $n\lt 12$) and it does not work if $n=12$, $n=13$, or $n=14$, because then $n-3\lt 12$, so our inductive hypothesis does not guarantee that we can make an $n-3$-cent stamp (the proposition "works" for any $n\lt 12$ by default). So the argument above is not complete. We still need to make sure everything works for $n\lt 12$, $n=12$, $n=13$, and $n=14$. By "everything works", we mean "if the proposition is true for all $k$ strictly smaller than $n$, then it holds for $n$.
If $n\lt 12$, then this is true simply because the proposition is true for $n$, so the consequent is true.
If $n=12$, this is true because we can verify that we can make a $12$-cent stamp (four $3$-cent stamps). So the implication is true because the consequent is true.
If $n=13$, the implication is true because the consequent is true: we can make a $13$-cent stamp (a $7$-cent stamp and two $3$-cent stamps).
If $n=14$, the implication is true because the consequent is true: we can make a $14-$cent stamp (two $7$-cent stamps).
And if $n\geq 15$, the argument we had before already worked.
So now we have established the strong inductive step for every positive integer $n$, and so by strong induction we have established the desired proposition for all positive integers.
(The $3+n-2$ came from applying the inductive argument to $n+1$).
Personally, I prefer to do proofs by strong induction by first doing the "general case", and then doing the "special cases", as the latter are only revealed after we examine the general proof and see if it works for all $n$ or not. This also helps draw the distinction between proofs by strong induction and proofs by regular induction, specifically that the latter need a base and an inductive step, while the former only needs an inductive step (but may require special cases).
Added. See also this previous question
Best Answer
Actually, the proof for the first question doesn't really use strong induction, but uses what is sometimes called 'strong weak induction': you make reference to a fixed number of earlier cases (in this case, two) and you know exactly what those previous cases are (in this case, the two previous numbers). In fact, it is those latter cases that determine how many base cases there should be. For example, if your recursive relation was $u_{n+4} = 4u_{n+2} - 2u_n$, then it is clear you need $4$ base cases, for only after those first $4$ can you get the recursion up and running.
The proof for the second question does use a genuine case of strong induction, since you prove $P_{k+1}$ on the basis of all of the $P_1$ through $P_k$ (or, in this case, $P_2$ through $P_k$). So note that there is no longer a fixed number of cases to which we refer back (as it changes for each $k$), and it is also no longer clear which specific cases you use; you simply assume they all hold as the inductive hypothesis.
Now, depending on how you look at it, strong induction can in fact be said to have no 'base' cases at all: you simply show that the claim holds for any $k$ if you assume it holds for all previous ones: do this for any $k$, and you're done! No separate base cases needed. In practice, though, the very first one is a bit of an exception, in that it has no previous ones (in your case, there is no relevant case smaller than $2$, since the claim is about all whole numbers greater than $1$). So this means that you need to prove the case $P_2$ on the basis of nothing, i.e. you simply need to prove the case $P_2$ in and of itself .... which is why this first case is often seen as 'the base case' for strong induction.