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Homesquare of sum

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# square of sum

The well-known formula for squaring a sum of two numbers or terms is

$\displaystyle(a\!+\!b)^{2}\;=\;a^{2}\!+\!2ab\!+\!b^{2}.$ | (1) |

It may be derived by multiplying the binomial $a\!+\!b$ by itself.

Similarly one can get the squaring formula for a sum of three summands:

$\displaystyle(a\!+\!b\!+\!c)^{2}\;=\;a^{2}\!+\!b^{2}\!+\!c^{2}\!+\!2bc\!+\!2ca% \!+\!2ab$ | (2) |

Its contents may be expressed as the

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:

$\left(\sum_{i}a_{i}\right)^{2}\;=\;\sum_{i}a_{i}^{2}+2\!\sum_{{i<j}}a_{i}a_{j}.$ |

This is true for any number of summands. The rule may be formulated also as

$\displaystyle(a\!+\!b\!+\!c+...)^{2}\;=\;(a)a+(2a\!+\!b)b+(2a\!+\!2b\!+\!c)c+...$ | (3) |

which in the case of four summands is

$\displaystyle(a\!+\!b\!+\!c\!+\!d)^{2}\;=\;(a)a+(2a\!+\!b)b+(2a\!+\!2b\!+\!c)c% +(2a\!+\!2b\!+\!2c\!+\!d)d.$ | (4) |

One can use the idea of (3) to find the square root of a polynomial, when one tries to arrange the polynomial into the form of the right hand side of (3).

Related:

SquareRootOfPolynomial, DifferenceOfSquares, HeronianMeanIsBetweenGeometricAndArithmeticMean, ContraharmonicMeansAndPythagoreanHypotenuses, CompletingTheSquare, TriangleInequalityOfComplexNumbers

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## Mathematics Subject Classification

30-00*no label found*26-00

*no label found*11-00

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new correction: Error in proof of Proposition 2 by alex2907

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new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

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new question: A trascendental number. by Ron Castillo

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new question: Banach lattice valued Bochner integrals by math ias