I have a few doubts about the properties of sigma notation, $\Sigma$ . My questions rely on factorization and solving equations with $\Sigma$.On account of the fact that my questions are correlated, I will keep them in this single post rather than different posts. It is also important to note that I have read other questions here, however, these did not help me.
First. Suppose that we have the following expression,
$\displaystyle\left(\sum_{j=0}^{N}y^{j}\left(\sum_{i=0}^{j}k^{i}\,i\,\displaystyle A_{i,j-i}\right)\right)^{2}+\left(\sum_{j=0}^{N}y^{j}\left(\sum_{i=0}^{j}k^{i}\,i\,\displaystyle B_{i,j-i}\right)\right)^{2}$.
Is it allowed to factorize the first term,$\sum_{j=0}^{N}y^{j}$, from both expressions?
That is to say,
$\displaystyle\left(\sum_{j=0}^{N}y^{j}\left(\sum_{i=0}^{j}k^{i}\,i\,\displaystyle A_{i,j-i}\right)\right)^{2}+\left(\sum_{j=0}^{N}y^{j}\left(\sum_{i=0}^{j}k^{i}\,i\,\displaystyle B_{i,j-i}\right)\right)^{2}$
$=\displaystyle\left(\sum_{j=0}^{N}y^{j}\right)^{2}\left[\left(\sum_{i=0}^{j}k^{i}\,i\,\displaystyle A_{i,j-i}\right)^{2}+\left(\sum_{i=0}^{j}k^{i}\,i\,\displaystyle B_{i,j-i}\right)^{2}\right]$
$=\displaystyle\left(\frac{y^{N+1}}{y-1}-\frac{1}{y-1}\right)^{2}\left[\left(\sum_{i=0}^{j}k^{i}\,i\,\displaystyle A_{i,j-i}\right)^{2}+\left(\sum_{i=0}^{j}k^{i}\,i\,\displaystyle B_{i,j-i}\right)^{2}\right]$?
Second. It is already well-known that $\displaystyle\left(\sum_{j=0}^{N}x_{j}\right)^2=\sum_{j=0}^{N}(x_{j})^{2}+\sum_{m<j}^{N}x_{j}\,x_{m}$.
Nevertheless, how should I proceed if the squared summation has a product of terms, i.e. $\displaystyle\left(\sum_{i=0}^{j}\,\,k^{i}\,\,i\,\,\displaystyle A_{i,j-i}\right)^{2}$?
Third. How may one solve the equation $\left(\sum_{i=0}^{j}k^{i}\,i\,\displaystyle A_{i,j-i}\right)^{2}+\left(\sum_{i=0}^{j}k^{i}\,i\,\displaystyle B_{i,j-i}\right)^{2}=0$ for $k$?
Thanks in advance
No, even without the overall square you cannot factor $$\displaystyle\left(\sum_{j=0}^{N}y^{j}\left(\sum_{i=0}^{j}k^{i}\,i\,\displaystyle A_{i,j-i}\right)\right)=\left(\sum_{j=0}^{N}y^{j}\right)\left(\sum_{i=0}^{j}k^{i}\,i\,\displaystyle A_{i,j-i}\right)$$ as each term in the $j$ sum multiplies a different $i$ sum. It would be like claiming $(ab+cd)=(a+c)(b+d)$