Given the sequence $a_n=3n+4$, need to find the number of elements which divisible by $4$ without reminder and without using induction.
I can see that $a_4=16$ is the first term which divisible by $4$, and $a_{124}=376$ is the last term.
so i guess i need the distance of $+12$ between each element for it to be valid (starting $b_1=16$ till $b_n=376$).
My problem is that it's all intuition.
How can i show all that in a mathematical writing?
Thanks.
EDIT: there are 126 elements in that sequence.
The number $3n+4$ is divisible by $4$ whenever $n$ is divisible by $4$. This can easily be seen by writing $n=4k$, so $a_n=3\cdot4k+4=4\cdot(3k+1)$. Since there are an infinite number of choices for $n$ of the form $4k$, there are clearly an infinite amount of values of $3k+4$ which are divisible by $4$.
EDIT: I think I understand your problem better now, it is just that you forgot to include the bounds to your sequence. It should be noted that if $3n+4$ is divisible by $4$, then $3n+4-4=3n$ is also divisible by $4$, and so $n$ must be divisible by $4$. Thus the entirety of the values in your sequence which are divisible by $4$ are those of the form $3\cdot 4k+4=12k+4$, and as you have guessed, the difference between two consecutive values of this form is $12$
EDIT 2: One way you can do this is to note that 1 out of every 4 numbers is divisible by four, so one way you can calculate how many numbers of the sequence are divisible by four is by finding $\lfloor{\frac{N}{4}}\rfloor$ where $N$ is the amount of numbers in the sequence. Since $N=126$, $\lfloor{\frac{N}{4}}\rfloor=\lfloor{31.5}\rfloor=31$