I believe the crux of Noble's question, as presented in his recent comment, is:
[B]ut I can't see how assuming it's true for more than one value is more powerful.
In logical terms, we say that a statement $A$ is stronger than a statement $B$ if
$A \implies B$. It is clear that -- forgive me for writing $\wedge$ for and when discussing logical statements --
$A \wedge A' \implies A$,
and more generally
$A_1 \wedge A_2 \wedge \ldots \wedge A_n \implies A_n$.
In other words, assuming a set of things is stronger than assuming a subset of things.
This is the sense in which strong induction is "stronger" than conventional induction: for your predicate $P$ indexed by the positive integers, assuming $P(1) \wedge \ldots \wedge P(n)$ is stronger than just assuming $P(n)$. In more practical terms, the more hypotheses you assume, the more you have to work with and it can only get easier to construct a proof.
Now let me supplement with further comments:
Nevertheless the principle of mathematical induction implies (and, more obviously, is implied by) the principle of strong induction, via the simple trick of switching from the predicate $P(n)$ to the predicate $Q(n) = P(1) \wedge \ldots P(n)$.
Here is a further possible source of confusion in the terminology. Suppose I have a theorem of the form $A \wedge B \implies C$. Someone else comes along and proves
the theorem $A \implies C$. Now their theorem is stronger than mine: i.e., it implies my theorem. Thus when you weaken the hypotheses of an implication you strengthen the implication. (While we're here, let's mention that if you strengthen the conclusion of an implication, you strengthen the implication.) This apparent reversal may be the locus of the OP's confusion.
With simple induction you use "if $p(k)$ is true then $p(k+1)$ is true" while in strong induction you use "if $p(i)$ is true for all $i$ less than or equal to $k$ then $p(k+1)$ is true", where $p(k)$ is some statement depending on the positive integer $k$.
They are NOT "identical" but they are equivalent.
It is easy to see that if strong induction is true then simple induction is true: if you know that statement $p(i)$ is true for all $i$ less than or equal to $k$, then you know that it is true, in particular, for $i=k$ and can use simple induction.
It is harder to prove, but still true, that if strong induction is true, then simple induction is true. That is what we mean by "equivalent".
Here we have a question. It is not why we still have "weak" induction - it's why we still have "strong" induction when this is not actually any stronger.
My opinion is that the reason this distinction remains is that it serves a pedagogical purpose. The first proofs by induction that we teach are usually things like $\forall n\left[\sum_{i=0}^n i= \frac{n(n+1)}{2}\right]$. The proofs of these naturally suggest "weak" induction, which students learn as a pattern to mimic.
Later, we teach more difficult proofs where that pattern no longer works. To give a name to the difference, we call the new pattern "strong induction" so that we can distinguish between the methods when presenting a proof in lecture. Then we can tell a student "try using strong induction", which is more helpful than just "try using induction".
Best Answer
The second line in the red box says $$a_{k+1}=10r2^k-12s2^{k-1}$$ We can write $10=2\cdot5$ and $12=2^2\cdot3$: $$a_{k+1}=2\cdot5r2^k-2^2\cdot3s2^{k-1}$$ Then we can collect the powers of two, yielding the third line in the red box: $$a_{k+1}=5r2^{k+1}-3s2^{k+1}$$
Weak induction just assumes $P(n)$ to be true when proving $P(n+1)$. Strong induction, on the other hand, uses all of $P(n),P(n-1),P(n-2),\dots$ down to the base cases, although in practice only a few of the highest proved propositions (but more than one) will be needed.