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EDIT: As per the comments, both are equivalent. I searched the top-voted questions under the tag [induction] but couldn’t find a Proof of the Second Principle, assuming the First Principle. The reverse (First from Second) is pretty easy, so I’d like to see a proof of Second from First.

I’m going through the first chapter (“Preliminaries”) of David M. Burton’s Elementary Number Theory, 7th Ed.

The First Principle of Finite Induction states that

Let S be a set of positive integers with the following properties:
(i) The integer $1$ belongs to S,
(ii) Whenever the integer $k$ is in S, the next integer $k+1$ must also be in S.
Then S is the set of all positive integers.

The Second Principle of Finite Induction states that

Let S be a set of positive integers with the following properties:
(i) The integer $1$ belongs to S,
(ii) Whenever the integers $1,2,…,k$ belong to S, the next integer $k+1$ must also be in S.
Then S is the set of all positive integers.

The author goes on to elaborate that both theorems can be generalised to start from any positive integer $n_0$ and shows us an example where the Second Principle (which is also called Strong Induction) is handier than the First:

Prove that for the nth term of the Lucas sequence $L_n$, $L_n< \left(\dfrac74\right)^n$.

Now my question is: Is there any technical reason (apart from “assuming all of $1,2,…,k$ is unnecessary when we can make do with only assuming $k$“) that the First Principle is preferred over the Second while doing things like:

Prove that $1+2+3+…+n=\dfrac{n(n+1)}{2}$ .

In other words, is the First Principle of Finite Induction essentially made redundant by the Second?

insipidintegrator
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    The principles are equivalent, so the distinction isn't important. – lulu Sep 06 '22 at 12:48
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    The Second Principle allows you to use a stronger assumption in your proof compared to the First principle. This may make your proof sometimes easier to write. For the example at the end of your question, you can write a proof using only the first principle. Technically, both principles are equivalent, i.e. each follows from the other. – Hans Engler Sep 06 '22 at 12:50
  • Let $T = {k \mid {1, \ldots, k} \subseteq S}$. $S = \mathbb N$ iff $T = \mathbb N$, and applying the second principle to $S$ is equivalent to applying the first principle to $T$. – Dmitry Sep 06 '22 at 15:30
  • @TrystwithFreedom No. i asked what’s the need for the First Principle? – insipidintegrator Sep 14 '22 at 17:01

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