Binomial ekspansion

Formlerne

Der findes grundlæggende tre binomialformler:

Vi kan forklare, hvorfor der findes 3 sådanne formler med en simpel udvidelse af produktet:

( a + b ) =2 ( a + b ) ⋅ ( a + b ) = a ⋅ a + a ⋅ b + b ⋅ a + b ⋅ a + b ⋅ b = a +22 ⋅ a ⋅ b + b {\displaystyle2 (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 -22 ⋅ a ⋅ a ⋅ b + b {{displaystyle2 (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 - a ⋅ b + b ⋅ a - b ⋅ b = a 2- b {\displaystyle2 (a+b)\cdot (a-b)=a\cdot a-a\cdot b+b\cdot a-b\cdot b=a^{2}-b^{2}}}

Ved hjælp af Pascals trekant

Hvis n {\displaystyle n} er et helt tal ( n ∈ Z {\displaystyle n\in \mathbb {Z} } ), bruger vi Pascals trekant.

For at udvide ( x + y ) {\displaystyle2 (x+y)^{2}} :

• finde række 2 i Pascals trekant (1, 2, 1)
• udvide x {\displaystyle x} og y {\displaystyle y}, således at x {\displaystyle x} effekt falder med 1 hver gang fra n {\displaystyle n} til 0, og y {\displaystyle y} effekt stiger med 1 hver gang fra 0 til n {\displaystyle n}
• gange tallene fra Pascals trekant med de rigtige udtryk.

Så ( x + y ) = 2x 1y2 +0 x2 y + x y 1+1 x 1y 0{\displaystyle2 (x+y)^{2}=1x^{2}y^{0}+2x^{1}y^{1}y^{1}+1x^{0}y^{2}}}

For eksempel:

( +3 x2 ) = ⋅2132 ⋅ ⋅ ( x2 ) + ⋅ 0231⋅ ⋅ ⋅ 1130⋅ ( x22 ) =2 + 9x12 + x + x 4{\displaystyle (3+2x)^{2}=1\cdot 3^{2}\cdot (2x)^{0}+2\cdot 3^{1}\cdot (2x)^{1}+1\cdot2 3^{0}\cdot (2x)^{2}=9+12x+4x^{2}}

Så som regel:

( x + y ) n = a x0 n y +0 a x1 n -1 y + 1a x2 n - 2y + 2⋯ + a n - 1x y 1n - 1+ a n x y 0n {\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}}

hvor a i {\displaystyle a_{i}}} er tallet i række n {\displaystyle n} og position i {\displaystyle i} i Pascals trekant.

Eksempler

( +5 x3 ) = ⋅3153 ⋅ ⋅ ( x3 ) + ⋅ 0352⋅ ⋅ 1⋅ ⋅351 ( x33 ) + ⋅ 2150⋅ ⋅ ( x3 ) {\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}}}3

= + 12575⋅3 x +15 ⋅9 x + ⋅ x + 21⋅ 27x =3 + 125x225 + x + x 135+ 2x 27{\displaystyle3 =125+75\cdot 3x+15\cdot 9x^{2}+1\cdot 27x^{3}=125+225x+135x^{2}+27x^{3}}}

( 5- 3x ) = ⋅3153 ⋅ ( -3 x ) + ⋅ 0352⋅ ⋅ ( - 3x ) + ⋅ 1351⋅ ⋅ ( -3 x ) + ⋅ 2150⋅ ⋅ ( - 3x ) {\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}}}3

= +125 +75 ⋅ ( -3 x ) +15 ⋅ 9x +21 ⋅ x + ⋅ ( - 27x )3 = 125- 223x + x 1352-27 x {\displaystyle3 =125+75\cdot (-3x)+15\cdot 9x^{2}+1\cdot (-27x^{3}})=125-223x+135x^{2}-27x^{3}}}

( +7 x4 )2 = ⋅ 5175⋅ ( x4 ) 2+ ⋅0574 ⋅ ⋅ ( x4 ) 2+ ⋅ 11073⋅ ⋅ ( x4 )2 + ⋅ 21072⋅ ⋅ ( x4 ) 2+ ⋅ 3571⋅ ⋅ ( x 4) 2+ ⋅ ⋅ 4170⋅ ( x4 ) 2{\displaystyle (7+4x^{2})^{5}=1\cdot5 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}}}

= + 1680712005⋅4 x +23430 ⋅16 x + 4490⋅ 64x +635 ⋅256 x + ⋅ x + 81⋅ 1024x {\displaystyle =16807+12005\cdot 4x^{2}+3430\cdot 16x^{4}+490\cdot 64x^{6}+35\cdot 256x^{8}+1\cdot 1024x^{10}}}10

= +16807 x48020 +2 x + x 54880+ 4x31360 +6 x + x 8960+ 8x 1024{\displaystyle10 \,=16807+48020x^{2}+54880x^{4}+31360x^{6}+8960x^{8}+1024x^{10}}}