square of sumSquareOfSum2005-10-09 11:27:372009-01-01 10:06:23Topicsquare% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}The well-known \PMlinkescapetext{formula} for squaring a sum of two numbers or \PMlinkescapetext{terms} is
\begin{align}
(a\!+\!b)^2 = a^2\!+\!2ab\!+\!b^2.
\end{align}
It may be derived by multiplying the binomial $a\!+\!b$ by itself.
Similarly one can get the squaring \PMlinkescapetext{formula} for a sum of three summands:
\begin{align}
(a\!+\!b\!+\!c)^2 = a^2\!+\!b^2\!+\!c^2\!+\!2bc\!+\!2ca\!+\!2ab
\end{align}
Its contents may be expressed as the
\textbf{Rule.}\, The square of a sum is equal to the sum of the squares of all the summands plus the sum of all the double products of the summands in twos:
$$\left(\sum_ia_i\right)^2 \;=\; \sum_ia_i^2+2\!\sum_{i \neq j}a_ia_j.$$\\
This is true for any number of summands.\, The rule may be formulated also as
\begin{align}
(a\!+\!b\!+\!c+...)^2 =
(a)a+(2a\!+\!b)b+(2a\!+\!2b\!+\!c)c+...
\end{align}
which in the case of four summands is
\begin{align}
(a\!+\!b\!+\!c\!+\!d)^2 =
(a)a+(2a\!+\!b)b+(2a\!+\!2b\!+\!c)c+(2a\!+\!2b\!+\!2c\!+\!d)d.
\end{align}
One can use the idea of (3) to find the \PMlinkescapetext{{\em square root of a polynomial}}, when one tries to arrange the polynomial into the form of the right hand \PMlinkname{side}{Equation} of (3).