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?
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.
Note change in limits and swapping of summation indices $x,y$: $$\begin{align} \sum_{1\le y\le x\le n}f(x,y) &=\sum_{x=1}^n\sum_{y=1}^x \ f(x,y)\\ &=\sum_{y=1}^n\sum_{x=y}^n\ f(x,y)\end{align}$$
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.
