I don't understand how supposing that $P(k), k\geq 1$ for ordinary induction is different from $P(i), 1 \leq i \leq k, k\geq1$ for strong induction. Example from quora:
Let’s say you wanted to prove that every positive integer has a prime factorization $𝑝_1𝑝_2𝑝_3...𝑝_𝑚$.
Let 𝑃(𝑛) be the statement that an integer 𝑛 has a prime factorization. We’ll proceed by strong induction. The basis is pretty clear, so I’ll leave it out.
Next we’ll assume that 𝑃(1),𝑃(2),𝑃(3),...,𝑃(𝑘) are true. 𝑘+1 can either be prime or composite, and if it’s prime we’re done, so we’ll assume it’s composite. That means 𝑘+1 can be written as a product of two positive integers, i.e. 𝑘+1=𝑝𝑞, with $𝑝,𝑞∈ℤ^+$. We can write 1<𝑝<𝑘+1, and 1<𝑞<𝑘+1, which implies that 2≤𝑝≤𝑘 and 2≤𝑞≤𝑘.
Here is why we need strong induction: if we had simply supposed 𝑃(𝑛) was true for arbitrary 𝑛, we would be stuck. However, we supposed that 𝑃(𝑛) was true for every positive integer up to 𝑛=𝑘, so we have much more information to work with. Because we supposed this, we know that 𝑃(𝑝) and 𝑃(𝑞) are true, i.e. that 𝑝 and 𝑞 can be represented as a product of primes. We were able to reduce the problem down to a point where 𝑝 and 𝑞 were in a range, and since our inductive hypothesis in strong induction supposes that 𝑃(𝑛) is true for a range of values (rather than just one arbitrary 𝑛), we can now use it to prove the truth of 𝑃(𝑘+1).
Using ordinary induction, I'd say that $P(p)$ and $P(q)$ are true because $2≤𝑝≤𝑘$ and $2≤𝑞≤𝑘$ and $P(k), k\geq 1$. Why can I not use ordinary induction here?
Another example is the proof that McCarthy 91 function equals 91 for all positive integers less than or equal to 101. The property is $P(n)=M(101-n), n \geq 0$ and $M(n)$ is the McCarthy function. The author of the proof calculates the base case for $P(0)$, then does a supposition that $P(i), 0 \leq i \leq k, k \geq 0$. The use of strong induction is justified by the fact that we need the inductive hypothesis to hold for $k-10$, but I don't see why $P(n), n\geq0$ wouldn't hold for $n=k-10, k\geq11$, that is $n$ is at least 1, if ordinary induction was used.