TITLE: What's the benefit of using strong induction when it's replaceable by weak induction?
QUESTION [6 upvotes]: 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

REPLY [4 votes]: 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.