If the summation just sum every term i was thinking that for instance 1+2+3+4 = 4+3+2+1
so why this $$\sum\limits_{i=1}^{n} (3i)\ = \sum\limits_{i=n}^{1} (3i) $$ is not true ?
And how i can invert the limits of the summation without changing the result ?
As a matter of convention, indices rise. If you want to reverse the order of the terms, you need to write $\sum_{i=1}^na_i=\sum_{i=1}^na_{n+1-i}$.
To be consistent with the summation identity $\sum\limits_a^{b-1}+\sum\limits_b^c=\sum\limits_a^c$ when $a<b<c$
You need to set $\displaystyle \sum\limits_{i=M}^m x_i=-\sum\limits_{i=m+1}^{M-1} x_i$ for the case $m<M\ $ (and $0$ if $|M-m|\le 1$).
Have a look at this paper for a detailed approach.
As for integrals when bounds are reversed, a negative sign appears before the sum.
as others pointed out it is common (not convention) that indexes start from low values to high in that order, but when it is obvious it is undertood
$\sum\limits_{i=1}^{n} (3i)\ = \sum\limits_{i=n}^{1} (3i) $ is true, not sure why you think $3+6+9+\cdots +3n \neq 3n+\cdots+9+6+3$?
Just remember it is a shorthand, it is meaningless without defining what it means, you coud even define it with additional properties $\sum\limits_{k \text{ is prime} \le10000} k $ means $2+3+5+7+11+\cdots$ only primes less than 10000.
