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 $_1_2_3..._$.
Let () be the statement that an integer 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 (1),(2),(3),...,() are true. +1 can either be prime or composite, and if it’s prime we’re done, so we’ll assume it’s composite. That means +1 can be written as a product of two positive integers, i.e. +1=, with $,∈ℤ^+$. We can write 1<<+1, and 1<<+1, which implies that 2≤≤ and 2≤≤.
Here is why we need strong induction: if we had simply supposed () was true for arbitrary , we would be stuck. However, we supposed that () was true for every positive integer up to =, so we have much more information to work with. Because we supposed this, we know that () and () are true, i.e. that and can be represented as a product of primes. We were able to reduce the problem down to a point where and were in a range, and since our inductive hypothesis in strong induction supposes that () is true for a range of values (rather than just one arbitrary ), we can now use it to prove the truth of (+1).
Using ordinary induction, I'd say that $P(p)$ and $P(q)$ are true because $2≤≤$ and $2≤≤$ 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.
Ordinary or weak induction proves $Q(n)$ for all $n\ge1$ with a base step $n=1$ and an inductive step from $n=k$ to $n=k+1$.
Complete or strong induction considers the special case where $Q(n)$ denotes "$P(k)$ for all $k$ from $1$ to $n-1$ inclusive". If we try to prove $Q(n)$ for all $n\ge1$ by weak induction, the base step is vacuously true, and the inductive step is showing that, if $P(k)$ for all $k$ from $1$ to $n-1$ inclusive, then $P(n)$. If we can prove this statement, the weak induction on $Q$ succeeds, and we have also proven $P(n)$ for all $n\ge1$.
In other words, strong induction states this: if "$P(k)$ for all $k$ from $1$ to $n-1$" implies $P(n)$, then $P(n)$ for all $n\ge1$. Usually, the $n-1$ is called $n$ instead, so we need to prove "$P(k)$ for all $k$ from $1$ to $n$" implies $P(n+1)$.
Unlike weak induction, strong induction does not in general need a base step. However, in some cases the argument proving its inductive step has to consider small values of $n$ as special cases.