In a double summation, can the range of the outer sum depend on the index variable of the inner sum? Given this summation:$$\sum_{1\le y\le x\le 3}2x+y=\sum_{x=1}^3\sum_{y=1}^x2x+y=\sum_{y=1}^3\sum_{x=y}^32x+y$$Am I allowed to write the summation as:$$\sum_{y=1}^x\sum_{x=1}^32x+y\space\space?$$
Basically, my question is: In a double summation, can the range of the outer sum depend on the index variable of the inner sum?
 A: No, the index variable can only appear inside the sum that uses this index variable.
A sum such as $\displaystyle\sum_{x=1}^3 (2x+y)$ represents a quantity that does not depend on any object named $x$. This sum is the same as $\displaystyle\sum_{b=1}^3 (2b+y)$ or $(2+y)+(4+y)+(6+y)$. It depends on the value of $y$ because $y$ is not the sum index.
Edit: The rule is that
$$\sum_{x}\sum_{y} q(x, y) = \sum_{x}\left(\sum_{y} q(x, y)\right)$$
If we take your last expression it would be
$$\eqalign{\sum_{y=1}^x \sum_{x=1}^3 (2x+y) &= \sum_{y=1}^x \left(\sum_{x=1}^3
 (2x+y)\right)\cr
&= \sum_{y=1}^x \big((2+y)+(4+y)+(6+y)\big)\cr
&= \sum_{y=1}^x (12+3y)
}$$
But now we have an undefined quantity $x$ in the bounds of the sum. It is meaningless.
A: Note change in limits and swapping of summation indices $x,y$:
$$\begin{align}
\sum_{1\le y\le x\le n}f(x,y)
&=\sum_{x=1}^n\sum_{y=1}^x \ f(x,y)\\
&=\sum_{y=1}^n\sum_{x=y}^n\ f(x,y)\end{align}$$
A: The reason for:
$$38=\sum_{1\le y\le x\le 3}2x+y \,\,\,\not\!\!\!\!\implies\sum_{y=1}^x\sum_{x=1}^32x+y\space\space =\frac{27x+3x^2}{2}$$
is that both $x,y$ are ultimately constrained between $1$ and $3$ on the LHS, while on the RHS the variable $x$ is constrained between $1$ and $3$ (and as a rule the inner sum is calculated before the outer sum), however $y$ is constrained between $1$ and $x$ and $x$ is no longer constrained implying $y$ is not constrained, which contradict the constraint ($y\le 3$) of the LHS.
In general:
$$\sum_{1\le y\le x\le 3}2x+y \,\,\,\not\!\!\!\!\implies$$
$$\sum_{y=1}^x\sum_{x=1}^32x+y \ \ \ \ or \ \ \ \sum_{x=y}^3\sum_{y=1}^32x+y \ \ \ \ or \ \ \ \sum_{x=1}^3\sum_{y=1}^32x+y \ \ \ \ or \ \ \ \sum_{y=1}^3\sum_{x=1}^32x+y$$
Note: The same rules apply to the double integrals.
