Induction proof equivalence In Induction, we do the following: Check $P(1)$ is true, then show that if $P(k)$ is true, then $P(k+1)$ is also true. So we proceed to assuming $P(k)$ is true, then attempt to show $P(k+1)$ is true, as the inductive hypothesize. 
But are we allowed to say, assume $P(k-1)$ is true (this is my inductive hypothesize) and then show that $P(k)$ is true? If so, what is the (if any) advantage of doing this?
The same goes for Strong Induction.
 A: I assume here that you are doing induction on positive integers.
There is no difference between the two approaches. In the first one, you formally prove
$$\forall k\geq 1, P(k) \Rightarrow P(k+1)$$
whereas in the second one it is
$$\forall k\geq 2, P(k-1) \Rightarrow P(k)$$
Proving $P(1)$ and any one of the two statements above will prove $\forall k\geq 1, P(k)$ by induction. The same goes for the so-called strong induction!
A: Yes, there is an advantage to using $P(k - 1) \Rightarrow P(k)$, it comes from structural induction.
You are probably familiar with performing induction on integers, proving $P(k+1)$ under the assumption $P(k)$.  However, sometimes we wish to prove statements about countable objects like lists, graphs, multisets, matrices, etc, not just integers.
There are 2 forms you could take.  
Form 1
Prove proposition $P(X)$, assuming it is true for all $P(Y)$ where $Y$ is a smaller version of $X$.  For example, $X$ could be a list, and $Y$ could range over all lists that are 1 element smaller than $X$.
Form 2
Assuming proposition $P(Y)$, prove that $P(X)$ is true for all $X$ that are 1 step larger than $Y$.  For example, you could assume $P(Y)$ holds for some graph $Y$, and attempt to prove that $P(X)$ holds for all graphs $X$ that have 1 more node than graph $Y$.
Clearly form 1 is easier to work with.  Form 1 attempts to prove 1 statement from a large number of assumptions, whereas form 2 requires proving a large number of statements from 1 assumption.  
Furthermore, in form 1 you get 1 more assumption to work with (sometimes you need it), you can assume that the statement you wish to prove is not the base case.  So for integers:
Form 1
$$(k > 0) \land P(k - 1) \rightarrow P(k)$$
Form 2
$$P(k) \rightarrow P(k + 1)$$
Ok, not a big deal for integers because you only have a single "1 step larger" integer than $k$.  However you get $k > 0$ for free, whereas $k + 1 > 0$, while easy to state, is obnoxious to have to state when needed.  Also Form 1 prepares you better for induction on more interesting objects.
