Binomul lui Newton

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elena_romina
Euclid
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Joined: Sat Nov 15, 2008 12:15 pm

Binomul lui Newton

Post by elena_romina »

Sa se demonstreze egalitatile:
\( S_1= 1+C_n^1cos \alpha+C_n^2cos2\alpha+...+C_n^ncos n\alpha=2^ncos^n\frac {\alpha}{2}cos(\frac{n\alpha}{2}) \)
\( S_2=C_n^1sin\alpha+C_n^2sin2\alpha+...+C_n^n sin n\alpha=2^ncos^n\frac{\alpha}{2}sin(\frac{n\alpha}{2}) \)
Trebuie sa calculez intai \( S_1+i S_2 \), apoi sa folosesc formula lui Moivre
Am ajuns aici:\( S_1+iS_2=1+C_n^1(cos\alpha+isin \alpha)+C_n^2(cos\alpha+isin\alpha)^2+..+C_n^n(cos\alpha+isin\alpha)^n \)
As avea nevoie de un mic ajutor..nu stiu ce sa fac mai departe..:?
Multumesc mult!
Marius Mainea
Gauss
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Location: Gaesti (Dambovita)

Post by Marius Mainea »

Aplica formula lui Moivre in fiecare paranteza apoi egaleaza cu \( S_1+iS_2=1+C_n^1z+C_n^2z^2+...+C_n^nz^n=(1+z)^n \) si in sfarsit egaleaza partile reale respectiv imaginare .
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elena_romina
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Post by elena_romina »

Am reusit sa o fac pana la urma. Mi-a dat ca \( (1+cos\alpha+isin\alpha)^n=2^n cos^n\frac{\alpha}{2}(cos(\frac{n\alpha}{2})+isin(\frac{n\alpha}{2})) \), apoi daca egalez partile reale si cele imaginare, rezulta concluzia.
Va multumesc foarte mult! :wink:
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