proof by induction, two basic questions about the procedure I'm currently learning to prove something via induction. I have two questions: 

1) In the inductive step, assuming that the statement $P(k-1)$ holds, so that you have to show that $P(k)$ follows, how exactly has the induction hypothesis to be formulated? Like this: 

"The statement $P(k)$ holds for some $k\le n-1.$" ? I guess there should be no difference between this 'variant' and this one where you assume $P(k)$ and then have to show that 'P(k+1)' in the inductive step, because it is

$$\big(P(k)\Rightarrow P(k+1)\big)\;\iff \big(P(k-1)\Rightarrow P(k)\big),$$right?

2) In the induction base, I'm still not sure in which cases I have to do verify the statement for more than one n, (i.e. not only n=0, but also n=1) and so on.. can you explain me typical situations in which I have to verify the statements for several n's for the induction base?

Thank you for any help.
First of all I applaud you for the username.

I'll do an induction proof for you, and show you why I do everything. Hopefully this will answer all your questions.

Suppose we have the following:

$$
f(n) = \left\{\begin{array}{lr}
        n, &amp; \text{for } 0\leq n\leq 2\\
        3f(n-2)+2f(n-3), &amp; \text{for } n&gt;2\\
        \end{array}\right\}
$$

Let's prove $f(n)&lt;2^n$ for all $n\in\mathbb{N}$

Let's define a predicate.

For all $n\in\mathbb{N}$, let $P(n):f(n)&lt;2^n$

Pf of $\forall n, P(n)$.



Base case(s)

First of all, notice the when we recurse ($n&gt;2$), we have to "go back" at most $3$ numbers, hence the $n-3$. This should suggest that we have $3$ base cases.

As for which ones, well definitely we must prove $P(0)$, and after that, we prove 2 more, so we prove $P(1),P(2)$. So in the inductive step, when we "go back", we have these base cases to work with, so we are okay. If you only proved $P(0)$, then in the inductive step you would get stuck, because you don't know anything regarding $P(n-2)$.

Let $n=0$

Then $f(n)=0$, by definition

$&lt; 2^0 = 1$

$\therefore P(0)$ holds.

Now do $P(1)$ and $P(2)$



Induction Step:
For of all, since we proved $P(0-2)$, now we let $n&gt;2$, or equally $n\geq 3$, so we enter the induction step.

Let $n&gt;2$

Suppose $P(j)$ holds where $0\leq j\leq n-1, j\in\mathbb{N}$. [IH]

What does this say? This says let $j$ be an arbitrary natural number between $0$ and $n-1$, both inclusive. This says that suppose $P(0)\land P(1)\land\dots\land P(n-1)$ hold. This is how I write my IH, it's very clear. Moreover, you can also write this:

Suppose $P(j)$ holds where $0\leq j\lt n, j\in\mathbb{N}$. [IH]

We want to prove that $P(n)$ holds. It's much easier to prove $P(n)$ holds, rather then assuming $P(n)$ and proving $P(n+1)$.

$f(n)=3f(n-2)+2f(n-3)$, by definition of recursive step

$&lt;3\cdot 2^{n-2}+2\cdot 2^{n-3}$, by IH, since $0\leq n-3&lt;n-2&lt;n$

$=6\cdot 2^{n-3}+2\cdot 2^{n-3}$

$=8\cdot 2^{n-3}$

$=2^n$

$\therefore P(n)$ holds.



We have showed it holds for all $\mathbb{N}$, and so our proof is done. For proving how many base cases, think about how many steps the recursion "goes back", and proceed from there. For writing your induction hypothesis, it becomes much easier with a predicate and proving $P(n)$. If you can understand the way I wrote mine, just use that. Assume the predicate holds within that range, where $0$ is your first base case, and $n-1$ is what you assume it holds to. Then prove $P(n)$. Hope you found this answer useful.
Assume you want to prove that

$$\forall n\ge n_0 \;\;P_n $$

In the induction base, you check that the first  $P_{n_0} $ is true.

In the inductive step , you start like this :

Let $n\ge n_0$ such that $P_n $.
then you prove $P_{n+1} $.