Teorema del binomio

Las fórmulas

Existen básicamente tres fórmulas de expansión binomial:

Podemos explicar por qué existen estas 3 fórmulas con una simple expansión del producto:

( a + b ) 2 = ( a + b ) ⋅ ( a + b ) = a ⋅ a + a ⋅ b + b ⋅ a + b ⋅ b = a 2 + 2 ⋅ a ⋅ b + b 2 {\displaystyle (a+b)^{2}=(a+b)\cdot (a+b)=a\cdot a+a\cdot b+b\cdot a+b\cdot b=a^{2}+2\cdot a\cdot b+b^{2}}.

( a - b ) 2 = ( a - b ) ⋅ ( a - b ) = a ⋅ a - a ⋅ b - b ⋅ a + b ⋅ b = a 2 - 2 ⋅ a ⋅ b + b 2 {\displaystyle (a-b)^{2}=(a-b)\cdot (a-b)=a\cdot a-a\cdot b-b\cdot a+b\cdot b=a^{2}-2\cdot a\cdot b+b^{2}}.

( a + b ) ⋅ ( a - b ) = a ⋅ a - a ⋅ b + b ⋅ a - b ⋅ b = a 2 - b 2 {\cdot (a+b)\cdot (a-b)=a\cdot a-a\cdot b+b\cdot a-b\cdot b=a^{2}-b^{2}}.

Usando el triángulo de Pascal

Si n {\displaystyle n} es un número entero ( n ∈ Z {\displaystyle n\in \mathbb {Z} } ), utilizamos el triángulo de Pascal.

Para expandir ( x + y ) 2 {\displaystyle (x+y)^{2}}:

• encontrar la fila 2 del triángulo de Pascal (1, 2, 1)
• expandir x {\displaystyle x} e y {\displaystyle y} para que la potencia de x {\displaystyle x} baje en 1 cada vez desde n {\displaystyle n} hasta 0 y la potencia de y {\displaystyle y} suba en 1 cada vez desde 0 hasta n {\displaystyle n}
• veces los números del triángulo de Pascal con los términos correctos.

Entonces ( x + y ) 2 = 1 x 2 y 0 + 2 x 1 y 1 + 1 x 0 y 2 {\displaystyle (x+y)^{2}=1x^{2}y^{0}+2x^{1}y^{1}+1x^{0}y^{2}}

Por ejemplo:

( 3 + 2 x ) 2 = 1 ⋅ 3 2 ⋅ ( 2 x ) 0 + 2 ⋅ 3 1 ⋅ ( 2 x ) 1 + 1 ⋅ 3 0 ⋅ ( 2 x ) 2 = 9 + 12 x + 4 x 2 {\displaystyle (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}}

Así que, por regla general:

( x + y ) n = a 0 x n y 0 + a 1 x n - 1 y 1 + a 2 x n - 2 y 2 + ⋯ + a n - 1 x 1 y n - 1 + a n x 0 y n {\displaystyle (x+y)^{n}=a_{0}x^{n}y^{0}+a_{1}x^{n-1}y^{1}+a_{2}x^{n-2}y^{2}+\cdots +a_{n-1}x^{1}y^{n-1}+a_{n}x^{0}y^{n}}

donde a i {desde el estilo a_{i}} es el número en la fila n {desde el estilo n} y la posición i {desde el estilo i} en el triángulo de Pascal.

Ejemplos

( 5 + 3 x ) 3 = 1 ⋅ 5 3 ⋅ ( 3 x ) 0 + 3 ⋅ 5 2 ⋅ ( 3 x ) 1 + 3 ⋅ 5 1 ⋅ ( 3 x ) 2 + 1 ⋅ 5 0 ⋅ ( 3 x ) 3 {\displaystyle (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 ⋅ 3 x + 15 ⋅ 9 x 2 + 1 ⋅ 27 x 3 = 125 + 225 x + 135 x 2 + 27 x 3 {\displaystyle =125+75\cdot 3x+15\cdot 9x^{2}+1\cdot 27x^{3}=125+225x+135x^{2}+27x^{3}}

( 5 - 3 x ) 3 = 1 ⋅ 5 3 ⋅ ( - 3 x ) 0 + 3 ⋅ 5 2 ⋅ ( - 3 x ) 1 + 3 ⋅ 5 1 ⋅ ( - 3 x ) 2 + 1 ⋅ 5 0 ⋅ ( - 3 x ) 3 {\displaystyle (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 ⋅ ( - 3 x ) + 15 ⋅ 9 x 2 + 1 ⋅ ( - 27 x 3 ) = 125 - 223 x + 135 x 2 - 27 x 3 {\displaystyle =125+75\cdot (-3x)+15\cdot 9x^{2}+1\cdot (-27x^{3})=125-223x+135x^{2}-27x^{3}}

( 7 + 4 x 2 ) 5 = 1 ⋅ 7 5 ⋅ ( 4 x 2 ) 0 + 5 ⋅ 7 4 ⋅ ( 4 x 2 ) 1 + 10 ⋅ 7 3 ⋅ ( 4 x 2 ) 2 + 10 ⋅ 7 2 ⋅ ( 4 x 2 ) 3 + 5 ⋅ 7 1 ⋅ ( 4 x 2 ) 4 + 1 ⋅ 7 0 ⋅ ( 4 x 2 ) 5 {\displaystyle (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 ⋅ 4 x 2 + 3430 ⋅ 16 x 4 + 490 ⋅ 64 x 6 + 35 ⋅ 256 x 8 + 1 ⋅ 1024 x 10 {\displaystyle =16807+12005\cdot 4x^{2}+3430\cdot 16x^{4}+490\cdot 64x^{6}+35\cdot 256x^{8}+1\cdot 1024x^{10}}.

= 16807 + 48020 x 2 + 54880 x 4 + 31360 x 6 + 8960 x 8 + 1024 x 10 {\displaystyle \},=16807+48020x^{2}+54880x^{4}+31360x^{6}+8960x^{8}+1024x^{10}}