What if the step one is falsy when checking the initial condition
What next can we do. Next step I think no so what it is proper to do
example prove that $3^{n+2}+6^{n+2}$ for any natural number $n\ge0$ is divisible by 15
$15a=3^{n+2}+6^{n+2}$ where $a\in \mathbb Z $
$n_0=0$
$15|3^{0+2}+6^{0+2}$
and I get $ 15|45 $
My question don't have sense because I get to know to not confuse $15|45$ with $45|15$ I thought $15|45$ is the same as $15 \div 45$ that's why this was unclear and first step is not falsy but true because $45 \div 15$
To respond to your example: Recall that "$a$ divides $b$" means that $na = b$ for some whole number $n$. $3 \cdot 15 = 45$, so $15$ does divide $45$; the statement is therefore true when $n = 0$.
Regarding what you seem to be asking, however: If the base case is false, the statement you are trying to prove is false. A simple example is this: Suppose I asked you to prove that all swans are white. You go out to find some swans, and the first swan you find is black. You wouldn't be confused about what to do next - you'd just know that I was wrong.
On the other hand, sometimes the thing you're trying to prove is almost true - that is, it's true once $n$ gets big enough. For example, $2^n > 3$. For $n = 0$, this is false - $2^0 = 1 < 3$. But the induction step works: if $2^n > 3$, then $2^{n + 1} = 2 \cdot 2^n > 2 \cdot 3 > 3$. So once it starts being true, it'll always be true. That just means we have to use a different base case; in this case, we start with $n = 2$, because that's the first $n$ for which $2^n > 3$. The end result is that what we prove is not "For all $n$, $2^n > 3$", because that's false; instead, we show that "For all $n \geq 2$, $2^n > 3$".