Binomial expansion

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Binomial expansion uses an expression to make a series. It uses a bracket expression like $(x + y) n$ .

[change] Using Pascal's triangle

If $n$ is an integer ( $n \in \mathbb{Z}$ ), we use Pascal's triangle.

To expand $(x + y) 2$ :

find row 2 of Pascal's triangle (1, 2, 1)
expand $x$ and $y$ so the $x$ power goes down by 1 each time from $n$ to 0 and the $y$ power goes up by 1 each time from 0 up to $n$
times the numbers from Pascal's triangle with the right terms.

So $(x + y) 2 = 1 x 2 y 0 + 2 x 1 y 1 + 1 x 0 y 2$

For example:

$(3+2x)^2 = 1 \cdot 3^2 \cdot (2x)^0 + 2 \cdot 3^1 \cdot (2x)^1 + 1 \cdot 3^0 \cdot (2x)^2 = 9 + 12x + 4x^2$

So as a rule:

$(x+y)^n = a_0x^ny^0 + a_1x^{n-1}y^1+ a_2x^{n-2}y^2+ \cdots + a_{n-1}x^{1}y^{n-1}+ a_nx^0y^n$

where $a i$ is the number at row $a$ and position $i$ in Pascal's triangle.

[change] Examples

$(5+3x)^3 = 1 \cdot 5^3 \cdot (3x)^0 + 3 \cdot 5^2 \cdot (3x)^1 + 3 \cdot 5^1 \cdot (3x)^2 + 1 \cdot 5^0 \cdot (3x)^3$

$= 125 + 75 \cdot 3x + 15 \cdot 9x^2 + 1 \cdot 27x^3 = 125 + 223x + 135x^2 + 27x^3$

$(5-3x)^3 = 1 \cdot 5^3 \cdot (-3x)^0 + 3 \cdot 5^2 \cdot (-3x)^1 + 3 \cdot 5^1 \cdot (-3x)^2 + 1 \cdot 5^0 \cdot (-3x)^3$

$= 125 + 75 \cdot (-3x) + 15 \cdot 9x^2 + 1 \cdot (-27x^3) = 125 - 223x + 135x^2 - 27x^3$

$(7+4x^2)^5 = 1 \cdot 7^5 \cdot (4x^2)^0 + 5 \cdot 7^4 \cdot (4x^2)^1 + 10 \cdot 7^3 \cdot (4x^2)^2 + 10 \cdot 7^2 \cdot (4x^2)^3 + 5 \cdot 7^1 \cdot (4x^2)^4 + 1 \cdot 7^0 \cdot (4x^2)^5$