Given that $h=(5^2)^n-1$, how do I go about proving that $h$ is divisible by $24$ for all positive integers $n$, using induction, assuming $h/24$ is whole?
Here is my current method:
Assume that $P(n)=25^n-1=24m$, given that $m∈N$(positive integers)
$P(n+1)=25^{n+1}-1$
$P(n+1)=25^n25-1$
$P(n+1)=25(24m+1)-1$
$P(n+1)=25(24m)+24$
$P(n+1)=24(25m+1)$
Thus, given that $m∈N$, $25m+1$ must be a whole number, and thus for any $P(n)$, there must be a $P(n+1)$ divisible by 24. My question is whether or not this is a valid inductive proof?
There is a mistake$$P(n+1) = 25(24m \color{red}+1) -1$$
If you have reach $25(24m)-26$, it is easy to see that the first part is divisible by $24$, but what about $26$? You have to justify it (though you are not suppose to be able to do so).