Wzór dwumianowy Newtona

Dla $n\in {\mathbb{N}}_{+}$ i $a,b\in \mathbb{R}$ mamy:

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

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

Przykład 4. Zapisz ${\left(2+n\right)}^5$w postaci sumy algebraicznej.

${\left(2+n\right)}^5=\left(\genfrac{}{}{0ex}{}{5}{0}\right)2^5+\left(\genfrac{}{}{0ex}{}{5}{1}\right)2^{4}n+\left(\genfrac{}{}{0ex}{}{5}{2}\right)2^3{n}^2+\left(\genfrac{}{}{0ex}{}{5}{3}\right)2^2{n}^3+\left(\genfrac{}{}{0ex}{}{5}{4}\right)2^{}{n}^4+\left(\genfrac{}{}{0ex}{}{5}{5}\right)n^5=$

$=1\cdot 32+5\cdot 16n+10\cdot 8n^2+10\cdot 4n^3+5\cdot 2n^4+1\cdot n^5=32+80n+80n^2+40n^3+10n^4+n^5$

Obliczenia pomocnicze:

$\left(\genfrac{}{}{0ex}{}{5}{2}\right)=\left(\genfrac{}{}{0ex}{}{5}{3}\right)=\frac{5!}{2!\cdot 3!}=\frac{3!\cdot 4\cdot 5}{2\cdot 3!}=10$
$\left(\genfrac{}{}{0ex}{}{5}{1}\right)=\left(\genfrac{}{}{0ex}{}{5}{4}\right)=5$