If one wants to prove a theorem of the form "For every $x$ of type $T$ and every natural number $n$, we have $P(x, n)$" using mathematical induction, then there are two possibilities:
One could first of all fix an arbitrary $x$ of type $T$, and then prove "For every natural number $n$, $P(x, n)$" via induction. In the base case we would have to show "$P(x, 0)$", and in the induction step we would have to show that if $P(x, n)$, then $P(x, n+1)$.
One proves that the statement "For every $x$ of type $T$, $P(x, n)$" holds for all natural numbers $n$ via induction. The base case would be "For all $x$ of type $T$, $P(x, 0)$"; and in the induction step we would have to show "For all $x$ of type $T$, $P(x, n+1)$" assuming "For all $x$ of type $T$, $P(x, n)$".
Can one say that in general, one of the methods 1 or 2 is easier? Which method should one choose?
Method 2 should be more widely applicable as it allows you to show $P(x,n+1)$ from $P(y,n)$ where possibly $y\ne x$. Formally, if you are able to show $$\forall x (P(x,n)\to P(x,n+1)) $$ as needed for method 1, then you also automatically have $$(\forall x P(x,n))\to (\forall x P(x,n+1)) $$ as needed for method 2 (hence metjod 2 cannot be more difficult), but not vice versa (hence method 1 may not always work).