Binomium van Newton

Er zijn in principe drie binomiale uitbreidingsformules:

We kunnen verklaren waarom er zo'n 3 formules zijn met een eenvoudige uitbreiding van het product:

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

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

Als n een geheel getal is ( n ∈ Z {\displaystyle n} in \mathbb {Z}} ), gebruiken we de driehoek van Pascal.

Om ( x + y ) 2 {{2}}
uit te breiden:

• vind rij 2 van Pascal's driehoek (1, 2, 1)
• breid x {stijl x} en y {stijl y} uit, zodat het vermogen van x {stijl x} telkens met 1 omlaag gaat vanaf n {stijl n} tot 0 en het vermogen van y {stijl y} telkens met 1 omhoog gaat vanaf 0 tot n {stijl n}
• maal de getallen uit Pascal's driehoek met de juiste termen.

Dus ( 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}}

Bijvoorbeeld:

( 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}}

Dus als een regel:

( 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 {{n} (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}}

waarbij a i {{i}} het getal op rij n {{i}} en positie i {{i}} in de driehoek van Pascal is.

Voorbeelden

( 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 5^{3}}^ (3x)^{0}+3 5^{2}} (3x)^{1}+3 5^{1} (3x)^{2}+1 (3x)^{0} (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 ⋅ 5^{3} (-3x)^{0}+3 ⋅ 5^{2} (-3x)^{1}+3 ⋅ 5^{1} (-3x)^{2}+1 ⋅ 5^{0} (-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 7^{5} (4x^{2})^{0}+5 7^{4} (4x^{2})^{1}+10 7^{3} (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 ⋅ 4x^{2}+3430 ⋅ 16x^{4}+490 ⋅ 64x^{6}+35 ⋅ 256x^{8}}+1 ⋅ 1024x^{10}}

= 16807 + 48020 x 2 + 54880 x 4 + 31360 x 6 + 8960 x 8 + 1024 x 10 {\an58020x^{2}+54880x^{4}+31360x^{6}+8960x^{8}+1024x^{10}}