\require{AMSmath}

Dit is een reactie op vraag 87338

Re: Binonium van Newton

Dat is het helemaal :)

Ronald
Leerling bovenbouw havo-vwo - donderdag 27 december 2018

Antwoord

Ok... daar komt ie...

$
\begin{array}{l}
\sum\limits_{k = 0}^{10} {\left( {\begin{array}{*{20}c}
{10} \\
k \\
\end{array}} \right) \cdot \left( {\frac{1}{2}} \right)^{10} } = \\
\left( {\begin{array}{*{20}c}
{10} \\
0 \\
\end{array}} \right) \cdot \left( {\frac{1}{2}} \right)^{10} + \left( {\begin{array}{*{20}c}
{10} \\
1 \\
\end{array}} \right) \cdot \left( {\frac{1}{2}} \right)^{10} + ... + \left( {\begin{array}{*{20}c}
{10} \\
{10} \\
\end{array}} \right) \cdot \left( {\frac{1}{2}} \right)^{10} \\
\left( {\frac{1}{2}} \right)^{10} \cdot \left( {\left( {\begin{array}{*{20}c}
{10} \\
0 \\
\end{array}} \right) + \left( {\begin{array}{*{20}c}
{10} \\
1 \\
\end{array}} \right) + ... + \left( {\begin{array}{*{20}c}
{10} \\
{10} \\
\end{array}} \right)} \right) \\
\left( {\frac{1}{2}} \right)^{10} \cdot 2^{10} = 1 \\
\end{array}
$

Dat was niet echt een verrassing of toch?

WvR
donderdag 27 december 2018

Re: Re: Binonium van Newton

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