What's the benefit of using strong induction when it's replaceable by weak induction? 
Example of a proof of a theorem using weak(ordinary) induction


The two types of inductions have process of proving P(a) and "for all integers $n \ge b, P(n)$" as a result in common. 
For example, in the proof of the following question, we can use weak induction instead of strong induction, and using weak(ordinary) induction makes the proof simpler and shorter than the strong form of induction. So what's the benefit of using strong induction when it's replaceable by weak induction?


[EDIT]
As requested, here's the weak induction version of the question. Plus, I changed the $s{k_1}$ part in red in the weak induction after reading answers.
I now understand using weak induction doesn't prove the statement.

Source: Discrete Mathematics with Applications, Susanna S. Epp
 A: The other two answers are of course correct, but given your comments on Brian's answer, I will give a more down-to-earth response: in all likelihood, the proof you have in mind using weak induction is not correct. You should do as Git Gud says and spell out exactly what alternative proof you have in mind.
Why do I suspect you don't have a weak proof of the claim Epp proves? Because, as Brian illustrates, if you take $P(n)$ to be the statement $$s_n=5^n-1$$
you can't show $P(k+1)$ solely on the basis of $P(k)$. 
Why? Just try it. By definition, $$s_{k+1}=6s_k-5s_{k-1}$$
and by the hypothesis $P(k)$, we can substitute $5^k-1$ for $s_k$, so we have $$s_{k+1}=6(5^k-1)-5\color{blue}{s_{k-1}}$$
But what do we do with $s_{k-1}$? We don't have any hypothesis on it! Only if we use strong induction can we use the hypothesis $P(k-1)$ and thereby substitute in $5^{k-1}-1$ for $s_{k-1}$.
A: In the example that you’re discussing, you can not leave out proving $P(1)$. To see this, change the definition of the sequence slightly: $s_0=0$, $s_1=5$, and $s_k=6s_{k-1}-5s_{k-2}$ for $k\ge 2$. As before, let $P(n)$ be the formula $s_n=5^n-1$. Then $P(0)$ is true, and the induction step proceeds exactly as before to show that if $P(i)$ is true for all integers $i$ from $0$ through $k$, and $k\ge 1$, then $P(k+1)$ is true: there is nothing in that part of the argument that depends on the value of $s_1$. However, it is obviously not the case that $P(n)$ is true for all integers $n\ge 0$, since $P(1)$ is false.
The induction step requires knowing that $P(k)$ and $P(k-1)$ are true: without both of those, you cannot derive the truth of $P(k+1)$. And that means that to get $P(2)$, you need to know not only that $P(0)$ is true, but also that $P(1)$ is true. Thus, this argument really does use strong induction.
In fact the two forms of induction are logically equivalent, and every argument using strong induction can be converted to one that uses ordinary induction, but the conversion requires changing the proposition $P$. I can’t remember whether Epp proves the equivalence; if she does not, leave a question, and I’ll explain further.
In practice one simply uses whatever form of induction is convenient. I really wish that elementary texts didn’t make such a big deal of the difference: it’s not really very important, apart from the fact that so-called strong induction is in many ways a better introduction to more sophisticated forms of induction like structural and transfinite induction.
A: Strong Induction is more intuitive in settings where one does not know in advance for which value one will need the induction hypothesis.
Consider the claim: 

Every integer $n \ge 2$ is divisible by a prime number.

Using strong induction the proof is straightforward. It is true for $n=2$, as $2 \mid 2$ and $2$ is prime. 
Assume the statement true for $2 \le a \le n$. 
We show $n+1$ is divisible by a prime number. 
If $n+1$ is a prime number, then as $(n+1) \mid (n+1)$, the claim is proved.
If $n+1$ is not a prime number then there exists some proper divisor $a\mid (n+1)$, so $2 \le a \le n$. 
By induction hypothesis, we know that $a$ is divisible by a prime number $p$. 
Since $p \mid a $ and $a \mid (n+1)$ it follows that $p \mid (n+1)$ and the proof is complete. 
If you want to do this with weak induction you will have to change the statement you want to prove to something less intuitive. 
A: Often it's conceptually easier to think about an argument in terms of strong induction.  It is true that you can always recast the proof in terms of weak induction (though I don't exactly follow the comments you're making on the example in the question--what you have to do to make the argument use weak induction is change the statement you are proving by induction from "$P(k)$" to "$P(i)$ is true for all $i\leq k$"), but it is a bit simpler to use strong induction directly.  I could turn your question around and ask you: why should you ever use weak induction, when it's replaceable by strong induction?  It's largely a matter of taste, and which version leads to a more easily understood proof for the particular problem you are trying to solve.
Besides that, the concept of strong induction generalizes better than weak induction.  Strong induction works on any well-ordered set (or even more generally, any set equipped with a well-founded relation).  This generalized form of induction, called transfinite induction, is extremely important in set theory and also has applications in many other areas of mathematics, and cannot be reduced to weak induction the same way ordinary induction can.  (Well, there is sort of a "weak induction" version of transfinite induction given by splitting into successor and limit cases, but it is still rather different from ordinary weak induction because of the limit cases besides $0$.  And this doesn't work at all when doing induction on general well-founded relations.)
