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De driehoek van Pascal en het binomium van Newton

Wat is het verband tussen de driehoek van Pascal en het binomium van Newton?

maarte
Leerling bovenbouw havo-vwo - woensdag 19 november 2003

Antwoord

Driehoek van Pascal

                             1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1

Binomium van Newton

(a+b)0 = 1
(a+b)1 = 1.a+1.b
(a+b)2 = 1.a2 + 2.ab + 1.b2
(a+b)3 = 1.a3 + 3.a2b + 3.ab2 + 1.b3
(a+b)4 = 1.a4 + 4.a3b + 6.a2b2 + 4.ab3 + b4
(a+b)5 = 1.a5 + 5.a4b + 10.a3b2 + 10.a2b3 + 5.ab4 + 1.b5
(a+b)6 = 1.a6 + 6.a5b + 15.a4b2 + 20.a3b3 + 15.a2b4 + 6.ab5 + 1.b6

Samengevat

$
\begin{array}{c}
\left( {\begin{array}{*{20}c}
0 \\
0 \\
\end{array}} \right) \\
\begin{array}{*{20}c}
{\left( {\begin{array}{*{20}c}
1 \\
0 \\
\end{array}} \right)} & {\left( {\begin{array}{*{20}c}
1 \\
1 \\
\end{array}} \right)} \\
\end{array} \\
\begin{array}{*{20}c}
{\left( {\begin{array}{*{20}c}
2 \\
0 \\
\end{array}} \right)} & {\left( {\begin{array}{*{20}c}
2 \\
1 \\
\end{array}} \right)} & {\left( {\begin{array}{*{20}c}
2 \\
2 \\
\end{array}} \right)} \\
\end{array} \\
\begin{array}{*{20}c}
{\left( {\begin{array}{*{20}c}
3 \\
0 \\
\end{array}} \right)} & {\left( {\begin{array}{*{20}c}
3 \\
1 \\
\end{array}} \right)} & {\left( {\begin{array}{*{20}c}
3 \\
2 \\
\end{array}} \right)} & {\left( {\begin{array}{*{20}c}
3 \\
3 \\
\end{array}} \right)} \\
\end{array} \\
\begin{array}{*{20}c}
{\left( {\begin{array}{*{20}c}
4 \\
0 \\
\end{array}} \right)} & {\left( {\begin{array}{*{20}c}
4 \\
1 \\
\end{array}} \right)} & {\left( {\begin{array}{*{20}c}
4 \\
2 \\
\end{array}} \right)} & {\left( {\begin{array}{*{20}c}
4 \\
3 \\
\end{array}} \right)} & {\left( {\begin{array}{*{20}c}
4 \\
4 \\
\end{array}} \right)} \\
\end{array} \\
\end{array}
$


Over de driehoek van Pascal kan je meer lezen op volgende links:Mvg
Els
woensdag 19 november 2003
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