Short Answer:
The sum $\sum_{j\in J} a + b +c+\cdots $ is taken over the first term following the $\sum$ operator. That is, everything before the first plus or minus sign. In my example, the first term is $a$. If you want to include more things in the first term, use parentheses. For instance, if you want to sum over the sum of $a$ and $b$ (but not $c$, etc), you write $\sum_{j\in J} (a + b) +c+\cdots$
Also, suppose $(x_j)_{j\in J}=x_1,x_2,\ldots,x_j$ is a sequence with domain $J$. It has been pointed out in the comments by SebastianSchoennenbeck, if one wants to take the sum over all numbers $j$ such that the sequence $(x_j)_{j\in J}$ is defined, it is clearer to write $\sum_{j\in J} (x_j + C_1)$ than $\sum_{j} (x_j + C_1)$. Knuth's definition of the $\sum$ operator agrees, stating
Formally, we write $\sum _{P(j)} x_j$ as an abbreviation for the sum
of all terms $x_j$ such that $j$ is an integer satisfying a given
property $P(k)$. A 'property $P(k)$' is any statement
about $k$ that can be either true or false (Knuth, Concrete Mathematics, 2e, p.23).
Observe that $j\in J$ is a statement, while $j$ is not, thus only $\sum_{j\in J}$ is correct notation.
Now, i'll discuss each of your suggestions. Note $\#1=\#3\neq \#2$!!!
- $\sum_{j\in J} (x_j + C_1) + C_2$
This is the correct notation for taking the sum over the sum of $x_j + C_1$. The parenthesis indicates that (the sum of) both terms $x_j,C_1$ are included in the summand (ie the thing to be summed). Since the $\sum $ operator is linear, we have that:
\begin{align} \sum_{j\in J} (x_j + C_1) + C_2
&= \sum _{j\in J} x_j +\sum _{j\in J} C_1 +C_2 \\
&= \sum _{j\in J} x_j +|J|\cdot C_1 +C_2 \\
\end{align}
with the last equality because $C_1$ is a constant, and any constant (say $C_1$) plus itself $|J|$ times, = $|J| \cdot C_1$.
- $[\sum_{j\in J} x_j + C_1] + C_2$
This notation only takes the sum of $x_j$. This expression is not equal to the other two because here $C_1$ is not in first term ($x_j$) and thus not in the summand. Thus your middle term is only a $C_1$ term instead of a $|J|\cdot C_1$.
This gets to the heart of your question: Only the first term after the $\sum$ is considered to be in the summand (the first term being the stuff before any $+,-$). If we want to take the sum over $x$ terms we enclose them in parentheses.
Writing out some simple sums may help you understand the notation. Suppose we have a sequence $(a_j)=a_1,a_2,a_3$ (whose domain is $J$={1,2,3}), and a constant C.
$$\sum_{j\in J} a_j + C= a_1 + a_2 +a_3 + C$$
Observe that since $C$ is not in the first term, and there are no parentheses, it is not in the summand, and we do not sum over it. In contrast:
\begin{align}
\sum_{j\in J} (a_j + C) &=
(a_1+C) + (a_2+C) +(a_3 + C)\\
&= \sum_{j\in J} (a_j) + 3C \\
&= \sum_{j\in J} (a_j)+ \sum_{j\in J}(C)
\end{align}
Finally, note that that $\sum _{j\in J} (x_j)$ is just a real number (real numbers are closed under addition), call it $R_1$. Then you have $[\sum_{j\in J} x_j + C_1] + C_2=(R_1+C_1)+C_2$. Of course $R_1+C_2$ is also a real number say $R_2$, thus making these substitutions, you wrote $(R_2)+C_2$. Thus the parentheses are clearly redundant.
- $[\sum_{j\in J} (x_j + C_1)] + C_2$
This is just #1 with a redundant pair of outer parentheses. That is $\sum_{j\in J} (x_j + C_1)$ is just a real number say $R_3$. Thus you wrote $ (R_3)+C_1$.
References:
Stewart, Calculus, 7e, p, A34-A37.
Knuth, Concrete Mathematics, 2e, 21-33.
