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
Best Answer
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 $s_{k-1}$.