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\abstract{Indices Algebra, regardless the context about Tensor Calculus}\footnote{Compiled with the on-line service at: http://latex.informatik.uni-halle.de/latex-online/latex.php. As every draft for a translation from foreign languages, this one missed lot of text in Spanish by the author, but he hopes to deliver the main ideas and relevant considerations of the matters treated here. By: R. J. Cano, Jan 3 2013 13:56 VET}
\subsubsection{Developing useful identities:}
About $(A + B)^{2} = A^{2} + 2AB + B^2$. Motivation: It's needed when writing down the Lagrangian for a multiple plane pendulum (double, triple, ....)\\
\'x$_{{i}}^{2}=[\sum _{j=1}^{i}L_{{j}}\cos \left( \theta_{{j}} \right)\theta^{'}_{{j}}]^{2}$\newline
\'y$_{{i}}^{2}=[\sum _{j=1}^{i}L_{{j}}\sin \left( \theta_{{j}} \right)\theta^{'}_{{j}}]^{2}$\newline

Abstraction for the general term: $L_{{j}} F \left( \theta_{{j}} \right)\theta^{'}_{{j}}$ \\
$A_{j} = L_{{j}} F \left( \theta_{{j}} \right)$\\
$B_{j} = \theta^{'}_{{j}}$ \\
Now, additionally: \\
$F \left( \theta_{{j}} \right)$ = Def. as either sine or cosine depending on j. \\

The problem is: $\left(A_{1}B_{1} + A_{2}B_{2} + ... +A_{i-1}B_{i-1} + A_{i}B_{i} \right)^{2} = $ ???; then lets build a "multiplication table":
(Please see above at beginning)
\begin{table}
\begin{center}
\begin{tabular}[t]{||l||l|l|l|l|l|l||}
\hline\hline
%empieza
 & $A_{1}B_{1}$ & $A_{2}B_{2}$ & $A_{3}B_{3}$ & ... & $A_{(i-1)}B_{(i-1)}$ & $A_{i}B_{i}$ \\
\hline\hline
$A_{1}B_{1}$ & $A_{1}^{2}B_{1}^{2}$ & . & $A_{1}B_{1}A_{3}B_{3}$ & . & . & .  \\
\hline
$A_{2}B_{2}$ & . & $A_{2}^{2}B_{2}^{2}$ & . & . & . & .  \\
\hline
$A_{3}B_{3}$ & $A_{3}B_{3}A_{1}B_{1}$ & . & $A_{3}^{2}B_{3}^{2}$ & . & $A_{3}B_{3}A_{(i-1)}B_{(i-1)}$ & .  \\
\hline
... & . & . & . & *** & . & .  \\
\hline
$A_{(i-1)}B_{(i-1)}$ & . & . & $A_{(i-1)}B_{(i-1)}A_{3}B_{3}$ & . & $A_{(i-1)}^{2}B_{(i-1)}^{2}$ & . \\
\hline
$A_{i}B_{i}$ & . & . & . & . & . &  $A_{i}^{2}B_{i}^{2}$ \\
%termina
\hline\hline
\end{tabular}
\end{center}
\end{table}

We might observe that there is $i^{2}$ elements in total for such table: i cols by i rows, exactly i of them in the diagonal upon the fact that due the commutative property of the multiplication in the real numbers, viewed as matrix such table is symmetrical, a key fact underlying to the following result regarding the proper way of counting the distinct kind of objects present in the table: \newline

$i^{2} = i + 2($"something")\newline

Moreover, developing the product:\newline

$(\sum _{j=1}^{i}A_{j}B_{j})^{2} = $\newline\newline\newline
$\sum _{j=1}^{i}A_{j}^{2}B_{j}^{2}$\newline\newline
$ + 2(A_{1}B_{1})(A_{2}B_{2}) + 2(A_{1}B_{1})(A_{3}B_{3}) + ... + 2(A_{1}B_{1})(A_{(i-1)}B_{(i-1)}) + 2(A_{1}B_{1})(A_{i}B_{i})$\newline
$ + 2(A_{2}B_{2})(A_{3}B_{3}) + ... + 2(A_{2}B_{2})(A_{(i-1)}B_{(i-1)}) + 2(A_{2}B_{2})(A_{i}B_{i})$\newline
$ + 2(A_{3}B_{3})(A_{(i-1)}B_{(i-1)}) + 2(A_{3}B_{3})(A_{i}B_{i})$\newline
.\newline
.\newline
.\newline
$ + 2(A_{(i-2)}B_{(i-2)})(A_{(i-1)}B_{(i-1)}) + 2(A_{(i-2)}B_{(i-2)})(A_{(i)}B_{(i)})$\newline
$ + 2(A_{(i-1)}B_{(i-1)})(A_{i}B_{i})$\newline

What doesn't appears over the diagonal, belongs to a multiple sum of the kind: $2(A_{k}B_{k})(A_{s}B_{s})$ where "k" runs from 1 until (i-1) and "s" runs from (k+1) until i.\newline\newline

Convinced of the previously described it should be true that:\newline

%\begin{table}
%\begin{center}
%\begin{tabular}[t]{|l|}
%\hline
$(\sum _{j=1}^{i}A_{j}B_{j})^{2} = \sum _{j=1}^{i}A_{j}^{2}B_{j}^{2} + 2(\sum_{k=1}^{(i-1)}\sum_{s=k+1}^{i}A_{k}B_{k}A_{s}B_{s})$ \\
%\hline
%\end{tabular}
%\end{center}
%\end{table}
\newline

Being this last expression necessarily a general identity in Algebra due to the following: Making i=2 and $B_{j}=1$ for all j, then we will get the well known elementary formula: \newline\newline

(i=2, $B_{j}=1$)\\
$(\sum _{j=1}^{2}A_{j}B_{j})^{2} = \sum _{j=1}^{2}A_{j}^{2}B_{j}^{2} + 2(\sum_{k=1}^{(1)}\sum_{s=2}^{2}A_{k}B_{k}A_{s}B_{s})$ \\
$[\sum _{j=1}^{2}A_{j}(1)]^{2} = \sum _{j=1}^{2}A_{j}^{2}(1)^{2} + 2[\sum_{k=1}^{(1)}\sum_{s=2}^{2}A_{k}A_{s}(1)^{2}]$ \\
$(\sum _{j=1}^{2}A_{j})^{2} = \sum _{j=1}^{2}A_{j}^{2} + 2A_{1}A_{2}$ \\ \\
$(A_{1} + A_{2})^{2} = A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2}$ \newpage

And also as sub-consequence: \newline

\begin{center}
$i^{2} = i + 2($"Something") \newline
\end{center}

Which would become this other:

$\\$
$1^{st}$ form: $N^{2} = N + 2\sum_{k=1}^{(N-1)}\sum_{s=k+1}^{N} $\\ \\
$2^{nd}$ form: $N^{2} = N + 2\sum_{k=1}^{(N-1)}$ \\ \\
$3^{rd}$ form: $N^{2} = \sum_{k=1}^{(N-1)} + \sum_{k=1}^{N}$ \\ \\

Note: Viewed as operators, the summation sigmas should be understood such that in the absence of operands (namely "general terms") at their right, they are operating on the unit (the possibly called: "trivial operand", or assumed "ad priori").
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