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 1Binomium 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.b6Samengevat$\begin{array}{c} \left( {\begin{array}{*{20}c} 0 \\ 0 \\\end{array}} \right) \\ \begin{array}{*{20}c} {\left( {\begin{array}{*{20}c} 1 \\ 0 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 1 \\ 1 \\\end{array}} \right)} \\\end{array} \\ \begin{array}{*{20}c} {\left( {\begin{array}{*{20}c} 2 \\ 0 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 2 \\ 1 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 2 \\ 2 \\\end{array}} \right)} \\\end{array} \\ \begin{array}{*{20}c} {\left( {\begin{array}{*{20}c} 3 \\ 0 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 3 \\ 1 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 3 \\ 2 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 3 \\ 3 \\\end{array}} \right)} \\\end{array} \\ \begin{array}{*{20}c} {\left( {\begin{array}{*{20}c} 4 \\ 0 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 4 \\ 1 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 4 \\ 2 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 4 \\ 3 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 4 \\ 4 \\\end{array}} \right)} \\\end{array} \\ \end{array}$Over de driehoek van Pascal kan je meer lezen op volgende links:Binomium van Newton en de driehoek van PascalDriehoek van PascalDriehoek van Pascal bij WisfaqMvg Els woensdag 19 november 2003 ©2004-2024 WisFaq
Wat is het verband tussen de driehoek van Pascal en het binomium van Newton? maarte Leerling bovenbouw havo-vwo - woensdag 19 november 2003
maarte Leerling bovenbouw havo-vwo - woensdag 19 november 2003
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 1Binomium 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.b6Samengevat$\begin{array}{c} \left( {\begin{array}{*{20}c} 0 \\ 0 \\\end{array}} \right) \\ \begin{array}{*{20}c} {\left( {\begin{array}{*{20}c} 1 \\ 0 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 1 \\ 1 \\\end{array}} \right)} \\\end{array} \\ \begin{array}{*{20}c} {\left( {\begin{array}{*{20}c} 2 \\ 0 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 2 \\ 1 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 2 \\ 2 \\\end{array}} \right)} \\\end{array} \\ \begin{array}{*{20}c} {\left( {\begin{array}{*{20}c} 3 \\ 0 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 3 \\ 1 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 3 \\ 2 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 3 \\ 3 \\\end{array}} \right)} \\\end{array} \\ \begin{array}{*{20}c} {\left( {\begin{array}{*{20}c} 4 \\ 0 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 4 \\ 1 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 4 \\ 2 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 4 \\ 3 \\\end{array}} \right)} & {\left( {\begin{array}{*{20}c} 4 \\ 4 \\\end{array}} \right)} \\\end{array} \\ \end{array}$Over de driehoek van Pascal kan je meer lezen op volgende links:Binomium van Newton en de driehoek van PascalDriehoek van PascalDriehoek van Pascal bij WisfaqMvg
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