I am studding a research paper in winch author presented a analytical model for set traversal and different cases of time complexity. I am not understanding the one point in the model that is related to summation formulation equation n# 18.
$\frac{m_1 + 1}{2}t + \frac{l_2 + 1}{2}t + .... + \frac{l_n + 1}{2}t$
$ = \frac{1}{2} \bigr(m_1 + n + \sum_{a=l_2}^{l_n} a\bigr) t$
Why $l_2 + l_3 + l_4 + ..... + l_n = \sum_{a=l_2}^{l_n} a$
why its not $\sum_{a=2}^{n} l_a$
On
Your doubts are reasonable and your proposal is correct. The formula (18) in the referred paper and some more show all the same kind of typos (or miscalculations).
We have \begin{align*} l_2 + l_3 + l_4 + ..... + l_n\color{blue}{=\sum_{a=2}^{n} l_a} \end{align*}
and not $l_2 + l_3 + l_4 + ..... + l_n = \sum_{a=l_2}^{l_n} a$, since $$\sum_{a=l_2}^{l_n} a=\sum_{l_2\leq a\leq l_n}a$$
Sloppy notation.
The author intended $\sum\limits_{a \in \{l_2, l_3, .... , l_n\}} a$ which would be the same thing as $\sum_{a=2}^n l_a$.
However I'd argue that $\sum\limits_{a=l_2}^{l_n} a$ is not actually wrong per se.
The notation for $\sum$ is $\sum\limits_{\text{some condition}} term = $ "the sum of all the $terms$ for which that condition is true".
(A practical example of this would be $\sum\limits_{a|24;a\in \mathbb N} a = $ the sum of all the divisors of $24=1 + 2 + 3 + 4+ 6 + 8 +12 + 24$.)
This means the familiar $\sum\limits_{a=2}^n l_a$ actually means $\sum\limits_{a\in \{2,3,4,...n\}} l_a$ and that $_{a=2}^n$ is just shorthand for $a\in \{2.....n\}$.
Except when were we ever actually TAUGHT that $_{a=2}^n$ is just shorthand for $a \in \{2.....n\}$? Well, maybe you were but I never was. I had to pick it up on the streets...
Anyway, if $_{a=2}^n$ is proper shorthand for $a \in \{2,.... n\}$ then isn't it okay that $_{a=l_2}^{l_n}$ proper shorthand for $a \in \{l_2, .... l_n\}$?
Meh... maybe. I admit it looks weird and we have to think but... is it any more weird than $_{a=2}^n$?
(That was rhetorical.)
(Anyway, welcome to learning mathematics on the street.)
====ADDENDUM====
In a comment in Marcus Scheurer's answer, I realize that $\sum\limits_{a=l_2}^{l_n} a$ could just as legitimately be interpretated as $l_2 + (l_2 + 1) + (l_2 + 3) + ..... +(l_2 + \lfloor l_n - l_2\rfloor)$.
It would take a bit bit of a bean-counter mind to interpret it that way if the individual $l_i$ values weren't specifically spelled out.
But that is a technical and correct interpretation.
So... was the author wrong or right? Arguments can be made both ways.
But it's what the author did. And the author clearly meant it to be interpreted as $\sum\limits_{i=2}^n l_i$.
Those are the existential facts. What we do with them is up to us.