Ņūtona binoms

Binomiālā izvēršana izmanto izteiksmi, lai izveidotu virkni. Tā izmanto šādu iekavās izteiktu izteiksmi: ( x + y ) n {\displaystyle (x+y)^{n}}. {\displaystyle (x+y)^{n}}. Ir trīs binomiālie izvērsumi.

Formulas

Pamatā ir trīs binomiskās izvēršanas formulas:

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

  

1. (Plus)

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

2. (mīnus)

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

3. vieta (Plus mīnuss)

Mēs varam izskaidrot, kāpēc ir šādas 3 formulas, izmantojot vienkāršu reizinājuma paplašinājumu:

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

Izmantojot Paskala trīsstūri

Ja n {\displaystyle n}n ir vesels skaitlis ( n Z {\displaystyle n\in \mathbb {Z} }{\displaystyle n\in \mathbb {Z} } ), mēs izmantojam Paskāla trīsstūri.


Lai paplašinātu ( x + y ) 2 {\displaystyle (x+y)^{2}}{\displaystyle (x+y)^{2}} :

  • atrast Paskāla trīsstūra 2. rindu (1, 2, 1)
  • paplašiniet x {\displaystyle x}x un y {\displaystyle y}y , lai x {\displaystyle x}x jauda samazinātos par 1 katru reizi no n {\displaystyle n}n līdz 0 un y {\displaystyle y}y jauda palielinātos par 1 katru reizi no 0 līdz n {\displaystyle n}. n
  • reizina skaitļus no Paskāla trīsstūra ar pareizajiem locekļiem.


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


Piemēram:

( 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}}} {\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}}


Tātad parasti:

( 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}} {\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}}

kur a i {\displaystyle a_{i}}{\displaystyle a_{i}} ir skaitlis n {\displaystyle n}n rindā un i {\displaystyle i}{\displaystyle i} pozīcijā Paskāla trīsstūrī.

Piemēri

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


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