Editing Binomial theorem (section)

== Examples ==

The first few cases of the binomial theorem are:
<math display="block">\begin{align}
(x+y)^0 & = 1, \\[8pt]
(x+y)^1 & = x + y, \\[8pt]
(x+y)^2 & = x^2 + 2xy + y^2, \\[8pt]
(x+y)^3 & = x^3 + 3x^2y + 3xy^2 + y^3, \\[8pt]
(x+y)^4 & = x^4  + 4x^3y + 6x^2y^2 + 4xy^3 + y^4,
\end{align}</math>
In general, for the expansion of {{math|(''x'' + ''y'')<sup>''n''</sup>}} on the right side in the {{mvar|n}}th row (numbered so that the top row is the 0th row):
* the exponents of {{mvar|x}} in the terms are {{math|''n'', ''n'' − 1, ..., 2, 1, 0}} (the last term implicitly contains {{math|1=''x''<sup>0</sup> = 1}});
* the exponents of {{mvar|y}} in the terms are {{math|0, 1, 2, ..., ''n'' − 1, ''n''}} (the first term implicitly contains {{math|1=''y''<sup>0</sup> = 1}});
* the coefficients form the {{mvar|n}}th row of Pascal's triangle;
* before combining like terms, there are {{math|2<sup>''n''</sup>}} terms {{math|''x''<sup>''i''</sup>''y''<sup>''j''</sup>}} in the expansion (not shown); 
* after combining like terms, there are {{math|''n'' + 1}} terms, and their coefficients sum to {{math|2<sup>''n''</sup>}}.
An example illustrating the last two points: <math display="block">\begin{align}
(x+y)^3 & = xxx + xxy + xyx + xyy + yxx + yxy + yyx + yyy & (2^3 \text{ terms}) \\
        & = x^3 + 3x^2y + 3xy^2 + y^3 & (3 + 1 \text{ terms})
\end{align}</math> with <math>1 + 3 + 3 + 1 = 2^3</math>.

A simple example with a specific positive value of {{math|''y''}}:
<math display="block">\begin{align}
(x+2)^3 &= x^3 + 3x^2(2) + 3x(2)^2 + 2^3 \\
&= x^3 + 6x^2 + 12x + 8.
\end{align}</math>

A simple example with a specific negative value of {{math|''y''}}:
<math display="block">\begin{align}
(x-2)^3 &= x^3 - 3x^2(2) + 3x(2)^2 - 2^3 \\
&= x^3 - 6x^2 + 12x - 8.
\end{align}</math>

=== Geometric explanation ===
[[File:binomial_theorem_visualisation.svg|thumb|300px|Visualisation of binomial expansion up to the 4th power]]
For positive values of {{mvar|a}} and {{mvar|b}}, the binomial theorem with {{math|1=''n'' = 2}} is the geometrically evident fact that a square of side {{math|''a'' + ''b''}} can be cut into a square of side {{mvar|a}}, a square of side {{mvar|b}}, and two rectangles with sides {{mvar|a}} and {{mvar|b}}. With {{math|1=''n'' = 3}}, the theorem states that a cube of side {{math|''a'' + ''b''}} can be cut into a cube of side {{mvar|a}}, a cube of side {{mvar|b}}, three {{math|''a'' × ''a'' × ''b''}} rectangular boxes, and three {{math|''a'' × ''b'' × ''b''}} rectangular boxes.

In [[calculus]], this picture also gives a geometric proof of the [[derivative]] <math>(x^n)'=nx^{n-1}:</math><ref name="barth2004">{{cite journal | last = Barth | first = Nils R.| title = Computing Cavalieri's Quadrature Formula by a Symmetry of the ''n''-Cube | doi = 10.2307/4145193 | jstor = 4145193 | journal = The American Mathematical Monthly | volume = 111| issue = 9| pages = 811–813 | date=2004}}</ref> if one sets <math>a=x</math> and <math>b=\Delta x,</math> interpreting {{mvar|b}} as an [[infinitesimal]] change in {{mvar|a}}, then this picture shows the infinitesimal change in the volume of an {{mvar|n}}-dimensional [[hypercube]], <math>(x+\Delta x)^n,</math> where the coefficient of the linear term (in <math>\Delta x</math>) is <math>nx^{n-1},</math> the area of the {{mvar|n}} faces, each of dimension {{math|''n'' &minus; 1}}:
<math display="block">(x+\Delta x)^n = x^n + nx^{n-1}\Delta x + \binom{n}{2}x^{n-2}(\Delta x)^2 + \cdots.</math>
Substituting this into the [[definition of the derivative]] via a [[difference quotient]] and taking limits means that the higher order terms, <math>(\Delta x)^2</math> and higher, become negligible, and yields the formula <math>(x^n)'=nx^{n-1},</math> interpreted as
:"the infinitesimal rate of change in volume of an {{mvar|n}}-cube as side length varies is the area of {{mvar|n}} of its {{math|(''n'' &minus; 1)}}-dimensional faces".
If one integrates this picture, which corresponds to applying the [[fundamental theorem of calculus]], one obtains [[Cavalieri's quadrature formula]], the integral <math>\textstyle{\int x^{n-1}\,dx = \tfrac{1}{n} x^n}</math> – see [[Cavalieri's quadrature formula#Proof|proof of Cavalieri's quadrature formula]] for details.<ref name="barth2004" />

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