Re: Binonium van Newton

Dit is een reactie op vraag 87338

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?

donderdag 27 december 2018

Re: Re: Binonium van Newton

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