Why is it that $1+(1+2+3+4+5+6+\ldots+n)$, basically a triangular number plus $1$, doesn't divide by $3$ or $5$? As a self learner I am currently learning about triangular numbers, for which the formula is:
$$T(n)=\frac{n(n+1)}{2}$$
While playing with my calculator, I added 1 to each resulted number, and I noticed that none of the results divides by $3$ or $5$
I am assuming that I am correct about this observation, other wise please let me know.
So I have multiplied both sides of the original formula by $2$, and now I have:
$2(T(n))=n(n+1) = n^2+n$
Now my question still remains still wide open:
why is it that $(n(n+1)+1) \bmod 3 \neq 0$ and $(n(n+1)+1) \bmod 5 \neq 0$?
Or you can see it as  why $(n^2+n+1) \bmod 3 \neq 0$ and $(n^2+n+1) \bmod 5 \neq 0$?
Or you can see it as why $(1+2+3+4+5+6+\ldots+n)+1$, Basically why a triangular number plus $1$ does not dividing by $3$ nor $5$?
I have tried Google and also tried searching over here, but either I don't know what I am searching for or I simply can't find an answer.
I am trying to pull my head for a possible answer, but I just don't have a clue where to begin with.
Any answers or hints are appreciated. Also if this is a duplicated, I honestly couldn't find it, so please just close and refer me to it.

Another way to look at it is, starting from 0
first you add 1 mod 3, so that the first triangular number is 1 mod 3,
then you add 2 mod 3, which gives 0 mod 3,
then you add 0 mod 3, so that it is still 0 mod 3,
repeat indefinitely
So the triangular numbers mod 3 are: 1, 0, 0, 1, 0, 0, ...
Since they are never 2, adding one to a triangular number gives a number not divisible by 3.
You can make a similar argument for 5.
