"MATEMĀTIKA 11. KLASEI"
Par Ņūtona binoma formulu sauc
$\begin{array}{l}{\left(a+b\right)}^{n}=\\ {C}_{n}^{0}{a}^{n}{b}^{0}+{C}_{n}^{1}{a}^{n-1}{b}^{1}+{C}_{n}^{2}{a}^{n-2}{b}^{2}+{C}_{n}^{3}{a}^{n-3}{b}^{3}+...+{C}_{n}^{k}{a}^{n-k}{b}^{k}+...+{C}_{n}^{n-1}{a}^{1}{b}^{n-1}+{C}_{a}^{n}{a}^{0}{b}^{n}\end{array}$
Formulas labo pusi sauc par pakāpes izvirzījumu.
${C}_{n}^{k}$ sauc par binominālajiem koeficientiem.

Binominālie koeficienti ir tie paši skaitļi, kuri veido Paskāla trijstūri.
Binominālo koeficientu summa ir ${2}^{n}$.

$\begin{array}{l}{\left(a+b\right)}^{0}=1\\ {\left(a+b\right)}^{1}=1\cdot a+1\cdot b\\ {\left(a+b\right)}^{2}=1{a}^{2}+2\mathit{ab}+\mathit{1}{b}^{2}\\ {\left(a+b\right)}^{3}=1{a}^{3}+\mathit{3}{a}^{2}b+\mathit{3}{\mathit{ab}}^{2}+\mathit{1}{b}^{3}\\ {\left(a+b\right)}^{4}=1{a}^{4}+4{a}^{3}b+\mathit{6}{a}^{2}{b}^{2}+4{\mathit{ab}}^{3}+\mathit{1}{b}^{4}\\ \end{array}$   $\begin{array}{l}{C}_{0}^{0}=1\\ {C}_{n}^{n}=1\\ {C}_{n}^{1}=n\end{array}$
Piemērs:
Izvirzīt pēc Ņūtona binoma formulas.
$\begin{array}{l}{\left(x-2y\right)}^{6}={\left(x+\left(-2y\right)\right)}^{6}=\\ \\ ={X}^{6}+{6x}^{5}\left(-2y\right)+15{x}^{4}{\left(-2y\right)}^{2}+20{x}^{3}{\left(-2y\right)}^{3}+15{x}^{2}{\left(-2y\right)}^{4}+6x{\left(-2y\right)}^{5}+{\left(-2y\right)}^{6}=\\ \\ ={x}^{6}-{12x}^{5}y+60{x}^{4}{y}^{2}-160{x}^{3}{y}^{3}+240{x}^{2}{y}^{4}-{192\mathit{xy}}^{5}+{64y}^{6}\end{array}$
Piemērs:
Aprēķināt ${\left(3a+b\right)}^{6}$ izvirzījuma vidējo locekli.

Risinājums:
Izvirzījumā ir $$6 + 1 = 7$$ locekļi, tātad vidējais ir 4. loceklis.

${T}_{4}={T}_{3+1}={C}_{6}^{3}{\left(3a\right)}^{6-3}\cdot {b}^{3}=\frac{6\cdot 5\cdot 4}{3\cdot 2\cdot 1}\cdot 27{a}^{3}{b}^{3}=540{a}^{3}{b}^{3}$
Piemērs:
Aprēķināt ${\left({x}^{2}+\frac{1}{x}\right)}^{12}$ izvirzījuma locekli, kas satur ${x}^{3}$.

Risinājums:
${T}_{k+1}={C}_{12}^{k}{\left({x}^{2}\right)}^{12-k}\cdot {\left({x}^{-1}\right)}^{k}={C}_{12}^{k}{x}^{24-2k}\cdot {x}^{-k}={C}_{12}^{k}{x}^{24-3k}$

Ja loceklis satur ${x}^{3}$, tad ${x}^{24-3k}={x}^{3}$ un $24-3k=3$, t.i. $k=\frac{24-3}{3}=7$.
Tātad tas ir $k+1=7+1=8.\phantom{\rule{0.147em}{0ex}}$ loceklis.

${T}_{8}={C}_{12}^{7}{x}^{3}={C}_{12}^{5}{x}^{3}=\frac{12\cdot 11\cdot 10\cdot 9\cdot 8}{5\cdot 4\cdot 3\cdot 2\cdot 1}\cdot {x}^{3}=\frac{95040}{120}\cdot {x}^{3}=792{x}^{3}$

Atbilde: 8. izvirzījuma loceklis ir $792{x}^{3}$.
"Binoms" no grieķu valodas tulkojumā nozīmē "divloceklis".
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