Question: I have found confronting information on other websites and thus will ask my question here.

Is the following conversion legitimate?

$$\sum\_{k=1}^n\sum\_{j=1}^nk\cdot j=\sum\_{k=1}^nk\cdot\sum\_{j=1}^nj$$

I have seen someone doing something similar on [Matroids Matheplanet](http://www.matheplanet.com/default3.html?call=viewtopic.php?topic=11746&ref=https%3A%2F%2Fwww.google.de%2F). 

Thank you for your help.

Philipp

Answer: 
Yes.

Since $k$ is a multiplier not related to $j$, it can be "factored" out, resulting in the product of two summations.

Simple example, using $n=3$:

$$\text{LHS}=\sum\_{k=1}^3\sum\_{j=1}^3 k\cdot j=\sum\_{k=1}^3 k+2k+3k=\sum\_{k=1}^3 6k=6\sum\_{k=1}^3 k=6(1+2+3)=36\\
\text{RHS}=\sum\_{k=1}^3k\sum\_{j=1}^3 j=(1+2+3)(1+2+3)=36$$

For the general case, 
$$\begin{align}
\text{LHS}&=\sum\_{k=1}^n\sum\_{j=1}^n k\cdot j\\
&=\sum\_{k=1}^n k+2k+3k+\cdots +nk\\
&=\sum\_{k=1}^3 (1+2+3+\cdots+n)k\\
&=\sum\_{k=1}^n \frac {n(n+1)}2k\\
&=\frac {n(n+1)}2\sum\_{k=1}^nk\\
&=\frac {n(n+1)}2\cdot\frac {n(n+1)}2\\
&=\sum\_{k=1}^n k\sum\_{j=1}^n j \end{align}$$

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*Note*: If $k$ is not a multiplier for the entire summand, then it cannot be "factored" out. 

Example:

$$\sum\_{k=1}^n\sum\_{j=1}^n j+k\neq \sum\_{k=1}^nk+\sum\_{j=1}^nj$$
> 
> Trivia: $\Large\sum, \int, \newcommand{\derivative}[3] {\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \derivative{n}{y}{x}$ are linear operations.
> 
> 
> 

What does it mean?

$\Large f(x)$ is linear if and only if

$$\Large f(x+y)=f(x)+f(y)$$

$$\Large f(ax)=a\cdot f(x)$$

Let 

$$\large f(x)=\sum\_{j=1}^nx$$

And

$$\large a=\sum\_{k=1}^nk$$

So

$$\large f(ax)= \sum\_{j=1}^n\left(\sum\_{k=1}^nk\right)x$$