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 = 1x2y0 + 2x1y1 + 1x0y2


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 ai 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
= 16807 + 12005 \cdot 4x^2 + 3430 \cdot 16x^4 + 490 \cdot 64x^6 + 35 \cdot 256x^8 + 1 \cdot 1024x^{10}
\, = 16807 + 48020x^2 + 54880x^4 + 31360x^6 + 8960x^8 + 1024x^{10}