Binomo formulė

Binomo formulė – dažnai dar vadinama Niutono formule, yra svarbi matematikos teorema, padedanti rasti dvinario, pakelto n-tuoju laipsniu, skleidinį. Teorema dažniausiai yra užrašoma

$(a+b{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a^{n-k}{b}^k$

arba

$(a+b{)}^n=(\genfrac{}{}{0ex}{}{n}{0})a^n+(\genfrac{}{}{0ex}{}{n}{1})a^{n-1}b+\dots +(\genfrac{}{}{0ex}{}{n}{k})a^{n-k}{b}^k+\dots +(\genfrac{}{}{0ex}{}{n}{n})b^n$

Skaičiai $(\genfrac{}{}{0ex}{}{n}{k})=\frac{n!}{k!\cdot (n-k)!}=C_n^k$ yra vadinami binomo koeficientais ir yra lygūs skaičiams iš atitinkamos Paskalio trikampio eilutės.

Arba

$(a+b{)}^n=C_n^0{a}^n{b}^0+C_n^1{a}^{n-1}{b}^1+C_n^2{a}^{n-2}{b}^2+C_n^3{a}^{n-3}{b}^3+\mathrm{...}+C_n^m{a}^{n-m}{b}^m+\mathrm{...}+C_n^{n-1}{a}^1{b}^{n-1}+C_n^n{a}^0{b}^n,$

kur $C_n^k$ yra deriniai. Jei $(a-b{)}^n$, tada bus tai minusas tai pliusas, pradedant nuo minuso, pvz:

$(a-b{)}^5=C_5^0{a}^5-C_5^1{a}^{4}b+C_5^2{a}^3{b}^2-C_5^3{a}^2{b}^3+C_5^{4}a{b}^4-C_5^5{b}^5$

Niutono formulė gali būti užrašyta dar taip:

$(a+b{)}^n=a^n+n{a}^{n-1}b+\frac{n(n-1)}{2!}{a}^{n-2}{b}^2+\frac{n(n-1)(n-2)}{3!}{a}^{n-3}{b}^3+\mathrm{...}+\frac{n(n-1)(n-2)\mathrm{...}(n-m+1)}{m!}{a}^{n-m}{b}^m+\mathrm{...}+na{b}^{n-1}+b^n.$

Skirtumo laipsnis:

$(a-b{)}^n=a^n-n{a}^{n-1}b+\frac{n(n-1)}{2!}{a}^{n-2}{b}^2-\frac{n(n-1)(n-2)}{3!}{a}^{n-3}{b}^3+\mathrm{...}+(-1{)}^m\frac{n(n-1)(n-2)\mathrm{...}(n-m+1)}{m!}{a}^{n-m}{b}^m+\mathrm{...}+(-1{)}^n{b}^n.$

• $(a+b{)}^2=a^2+2ab+b^2$
• $(a-b{)}^2=a^2-2ab+b^2$
• $(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$
• $(a-b{)}^3=a^3-3a^{2}b+3a{b}^2-b^3$

Pastaba, $0!=1.$

Penktos eilės Niutono binomo formulė yra tokia:

$(a-b{)}^5=C_5^0{a}^5-C_5^1{a}^{4}b+C_5^2{a}^3{b}^2-C_5^3{a}^2{b}^3+C_5^{4}a{b}^4-C_5^5{b}^5=$

$=\frac{5!}{0!\cdot (5-0)!}{a}^5-\frac{5!}{1!\cdot (5-1)!}{a}^{4}b+\frac{5!}{2!\cdot (5-2)!}{a}^3{b}^2-\frac{5!}{3!\cdot (5-3)!}{a}^2{b}^3+\frac{5!}{4!\cdot (5-4)!}a{b}^4-\frac{5!}{5!\cdot (5-5)!}{b}^5=$

$=\frac{5!}{1\cdot 5!}{a}^5-5a^{4}b+\frac{5!}{2!\cdot 3!}{a}^3{b}^2-\frac{5\cdot 4\cdot 3}{3\cdot 2}{a}^2{b}^3+\frac{5!}{4!}a{b}^4-\frac{5!}{5!\cdot 0!}{b}^5=$

$=a^5-5a^{4}b+\frac{5\cdot 4}{2}{a}^3{b}^2-5\cdot 2a^2{b}^3+5a{b}^4-b^5=$

$=a^5-5a^{4}b+10a^3{b}^2-10a^2{b}^3+5a{b}^4-b^5.$

Užrašysime ketvirto laipsnio Niutono binomo formulę:

$(a-b{)}^4=C_4^0{a}^4{b}^0-C_4^1{a}^{3}b+C_4^2{a}^2{b}^2-C_4^{3}a{b}^3+C_4^4{a}^0{b}^4=$

$=\frac{4!}{0!(4-0)!}{a}^4{b}^0-\frac{4!}{1!(4-1)!}{a}^{3}b+\frac{4!}{2!(4-2)!}{a}^2{b}^2-\frac{4!}{3!(4-3)!}a{b}^3+\frac{4!}{4!(4-4)!}{a}^0{b}^4=$

$=\frac{4!}{4!}{a}^4-\frac{4!}{3!}{a}^{3}b+\frac{4!}{2!\cdot 2!}{a}^2{b}^2-\frac{4!}{3!}a{b}^3+\frac{4!}{4!}{b}^4=$

$=a^4-4a^{3}b+\frac{4\cdot 3\cdot 2}{4}{a}^2{b}^2-4a{b}^3+b^4=$

$=a^4-4a^{3}b+6a^2{b}^2-4a{b}^3+b^4.$

Patikriname, kai a=6, b=2,

$(a-b{)}^4=(6-2{)}^4=4^4=256;$

$a^4-4a^{3}b+6a^2{b}^2-4a{b}^3+b^4=6^4-4\cdot 6^3\cdot 2+6\cdot 6^2\cdot 2^2-4\cdot 6\cdot 2^3+2^4=$

$=1296-4\cdot 216\cdot 2+6\cdot 36\cdot 4-4\cdot 6\cdot 8+16=1296-1728+864-192+16=256.$