Newtoni binoomvalem

Binoomi laiendamisel kasutatakse väljendit, et teha rida. See kasutab sulgudes väljendit nagu ( x + y ) n {\displaystyle (x+y)^{n}} $(x+y)^{n}$ . On olemas kolm binoomi laiendust.

Valemid

Põhimõtteliselt on olemas kolm binoomi laiendamisvalemit:

( a + b ) =2 a +2 a 2b + b {\displaystyle2 (a+b)^{2}=a^{2}+2ab+b^{2}}} $(a+b)^{2}=a^{2}+2ab+b^{2}$		1. (pluss)
( a - b ) =2 a 2-2 a b + b {\displaystyle2 (a-b)^{2}=a^{2}-2ab+b^{2}}} $(a-b)^{2}=a^{2}-2ab+b^{2}$		2. (miinus)
( a + b ) ⋅ ( a - b ) = a2 - b {\displaystyle2 (a+b)\cdot (a-b)=a^{2}-b^{2}}} $(a+b)\cdot (a-b)=a^{2}-b^{2}$		3. (pluss-miinus)

Me võime seletada, miks on olemas sellised 3 valemit lihtsa laienduse tootega:

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

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

( a + b ) ⋅ ( a - b ) = a ⋅ a - a ⋅ b + b ⋅ a - b ⋅ b = a2 - b {\displaystyle2 (a+b)\cdot (a-b)=a\cdot a-a\cdot b+b\cdot a-b\cdot b=a^{2}-b^{2}} $(a+b)\cdot (a-b)=a\cdot a-a\cdot b+b\cdot a-b\cdot b=a^{2}-b^{2}$

Pascali kolmnurga kasutamine

Kui n {\displaystyle n} on täisarv ( n ∈ Z {\displaystyle n\in \mathbb {Z} } $n\in \mathbb {Z}$ ), kasutame Pascali kolmnurka.

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

leida Pascali kolmnurga 2. rida (1, 2, 1)
laiendada x {\displaystyle x} ja y {\displaystyle y} nii, et x {\displaystyle x} võimsus väheneb 1 võrra iga kord alates n {\displaystyle n} kuni 0 ja y {\displaystyle y} võimsus suureneb 1 võrra iga kord alates 0 kuni n {\displaystyle n}.
korda arvud Pascali kolmnurgast koos õigete terminitega.

Seega ( x + y ) =2 x 1y 2+0 x2 y 1+1 x1 y 0{\displaystyle2 (x+y)^{2}=1x^{2}y^{0}+2x^{1}y^{1}+1x^{0}y^{2}}} $(x+y)^{2}=1x^{2}y^{0}+2x^{1}y^{1}+1x^{0}y^{2}$

Näiteks:

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

Nii et reeglina:

( x + y ) n = a x0 n y +0 a x1 n -1 y + 1a x2 n - 2y + 2⋯ + a n -1 x y1 n -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}} $(x+y)^{n}=a_{0}x^{n}y^{0}+a_{1}x^{n-1}y^{1}+a_{2}x^{n-2}y^{2}+\cdots +a_{n-1}x^{1}y^{n-1}+a_{n}x^{0}y^{n}$

kus a i {\displaystyle a_{i}} $a_{i}$ on arv reas n {\displaystyle n} ja positsioonis i {\displaystyle i} $i$ Pascali kolmnurgas.

Näited

( +5 x 3) = ⋅3153 ⋅ ( x 3) + ⋅ 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 $(5+3x)^{3}=1\cdot 5^{3}\cdot (3x)^{0}+3\cdot 5^{2}\cdot (3x)^{1}+3\cdot 5^{1}\cdot (3x)^{2}+1\cdot 5^{0}\cdot (3x)^{3}$

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

( 5-3 x ) = ⋅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 $(5-3x)^{3}=1\cdot 5^{3}\cdot (-3x)^{0}+3\cdot 5^{2}\cdot (-3x)^{1}+3\cdot 5^{1}\cdot (-3x)^{2}+1\cdot 5^{0}\cdot (-3x)^{3}$

= +12575 ⋅ ( -3 x ) + 15⋅ 9x +21 ⋅ ( - 27x )3 = 125- 223x + x1352 -27 x {\displaystyle3 =125+75\cdot (-3x)+15\cdot 9x^{2}+1\cdot (-27x^{3})=125-223x+135x^{2}-27x^{3}}} $=125+75\cdot (-3x)+15\cdot 9x^{2}+1\cdot (-27x^{3})=125-223x+135x^{2}-27x^{3}$

( +7 x 4)2 = ⋅5175 ⋅ ( x4 )2 + ⋅ 0574⋅ ( x4 )2 + ⋅ 11073⋅ ( x4 )2 + ⋅ 21072⋅ ( x4 )2 + ⋅ 3571⋅ ( x ) + ⋅ ⋅ ( 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}}} $(7+4x^{2})^{5}=1\cdot 7^{5}\cdot (4x^{2})^{0}+5\cdot 7^{4}\cdot (4x^{2})^{1}+10\cdot 7^{3}\cdot (4x^{2})^{2}+10\cdot 7^{2}\cdot (4x^{2})^{3}+5\cdot 7^{1}\cdot (4x^{2})^{4}+1\cdot 7^{0}\cdot (4x^{2})^{5}$

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

= +16807 x48020 + x +2 x 54880+ 4x 31360+ 6x8960 +8 x 1024{\displaystyle10 \,=16807+48020x^{2}+54880x^{4}+31360x^{6}+8960x^{8}+1024x^{10}}} $\,=16807+48020x^{2}+54880x^{4}+31360x^{6}+8960x^{8}+1024x^{10}$

Küsimused ja vastused

K: Mis on binoomiline laiendus?

V: Binoomi laiendamine on matemaatiline meetod, mille puhul kasutatakse väljendit, et luua rida, kasutades sulgudes väljendit (x+y)^n.

K: Milline on binomiaalpaisutuse põhikontseptsioon?

V: Binoomi laiendamise põhikontseptsioon on binoomväljendi potentsi laiendamine jadaks.

K: Mis on binoomiline avaldis?

V: Binoomiline avaldis on algebraline avaldis, mis sisaldab kahte terminit, mis on ühendatud pluss- või miinusmärgiga.

K: Milline on binoomi laiendamise valem?

V: Binoomi laiendamise valem on (x+y)^n, kus n on eksponent.

K: Kui palju on olemas binoomi laiendusi?

V: On olemas kolme tüüpi binoomi laiendusi.

K: Millised on kolm tüüpi binoomi laiendust?

V: Binoomi laienduste kolm liiki on - esimene binoomi laiendus, teine binoomi laiendus ja kolmas binoomi laiendus.

K: Kuidas on binoomiline laiendus kasulik matemaatilistes arvutustes?

V: Binoomi laiendamine on kasulik matemaatilistes arvutustes, sest see aitab lihtsustada keerulisi väljendeid ja lahendada keerulisi probleeme.