Binomická veta

Vzorce

V zásade existujú tri vzorce binomického rozšírenia:

Prečo existujú takéto 3 vzorce, môžeme vysvetliť jednoduchým rozšírením súčinu:

( a + b ) = 2( a + b ) ⋅ ( a + b ) = a ⋅ a + a ⋅ b + 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 ⋅ 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 + 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}}

Použitie Pascalovho trojuholníka

Ak n {\displaystyle n} je celé číslo ( n ∈ Z {\displaystyle n\v \mathbb {Z} } ), použijeme Pascalov trojuholník.

Rozšírenie ( x + y ) {\displaystyle2 (x+y)^{2}} :

• nájsť riadok 2 Pascalovho trojuholníka (1, 2, 1)
• rozšíriť x {\displaystyle x} a y {\displaystyle y} tak, aby výkon x {\displaystyle x} klesol o 1 vždy od n {\displaystyle n} do 0 a výkon y {\displaystyle y} stúpol o 1 vždy od 0 do n {\displaystyle n}
• vynásobte čísla z Pascalovho trojuholníka správnymi výrazmi.

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

Napríklad:

( + 3x2 ) = ⋅2132 ⋅ ( x2 ) + ⋅ 0231⋅ ( x2 ) + ⋅ 1130⋅ ( x 2) =2 + 9x 12+ 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}}

Takže spravidla:

( x + y ) n = a x 0n y +0 a x1 n -1 y + 1a x2 n - 2y + 2⋯ + a n - 1x y 1n - 1+ a n x y0 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}}

kde a i {\displaystyle a_{i}} je číslo v riadku n {\displaystyle n} a na pozícii i {\displaystyle i} v Pascalovom trojuholníku.

Príklady

( + 5x3 ) = ⋅3153 ⋅ ( x3 ) + ⋅ 0352⋅ ( x3 ) + ⋅ 1351⋅ ( x 3) + ⋅ 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⋅ 3x + 15⋅9 x + 21⋅ 27x =3 + 125x225 + 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 ⋅ ( - 3x ) + ⋅ 0352⋅ ( - 3x ) + ⋅ 1351⋅ ( - 3x ) + ⋅ 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

= +12575 ⋅ ( - 3x ) + 15⋅ 9x + 21⋅ ( - 27x ) 3=125 -223 x + x 1352-27 x {\displaystyle3 =125+75\cdot (-3x)+15\cdot 9x^{2}+1\cdot (-27x^{3})=125-223x+135x^{2}-27x^{3}}

( + 7x4 ) 2= ⋅5175 ⋅ ( x4 )2 + ⋅ 0574⋅ ( x4 )2 + ⋅ 11073⋅ ( x 4)2 + ⋅ 21072⋅ ( x 4)2 + ⋅ 3571⋅ ( x4 ) 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⋅64 x + 635⋅256 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 x54880 +4 x 31360+ x 68960+ 8x 1024{\displaystyle10 \,=16807+48020x^{2}+54880x^{4}+31360x^{6}+8960x^{8}+1024x^{10}}