Question

Combinari n luate 1+combinari de n luate4+combinari n luate 7+......+combinari n luate de n

Answer 1

$\displaystyle Fie~\varepsilon~o~radacina~nereala~de~ordinul~3~a~unitatii. \\ \\ \varepsilon^3=1 \Leftrightarrow (\varepsilon-1)(\varepsilon^2+\varepsilon+1)=0 \Leftrightarrow \varepsilon^2+ \varepsilon+1=0. \\ \\ Fie~P(x)=(1+x)^n=C^0_n+C^1_nx+C^2_nx^2+C^3_nx^3+...+C^n_nx^n. \\ \\ Avem:$
$\displaystyle P(1)=C^0_n+C^1_n+C^2_n+C^3_n+C^4_n+C^5_n+C^6_n+C^7_n+... \\ \\ P(\varepsilon)=C^0_n+C^1_n \varepsilon+C^2_n\varepsilon^2+C^3_n+C^4_n\varepsilon+C^5_n\varepsilon^2+C^6_n+C^7_n\varepsilon+... \\ \\ P(\varepsilon^2)=C^0_n+C^1_n\varepsilon^2+C^2_n\varepsilon+C^3_n+C^4_n\varepsilon^2+C^5_n\varepsilon+C^6_n+C^7_n\varepsilon^2+...$

$\displaystyle Deci \\ \\ P(1)=C^0_n+C^1_n+C^2_n+C^3_n+C^4_n+C^5_n+C^6_n+C^7_n+... \\ \\ \varepsilon^2P(\varepsilon)=C^0_n\varepsilon+C^1_n+C^2_n\varepsilon+C^3_n\varepsilon^2+C^4_n+C^5_n\varepsilon+C^6_n\varepsilon^2+C^7_n+... \\ \\ \varepsilon P(\varepsilon^2)=C^0_n\varepsilon+C^1_n+C^2_n\varepsilon^2+C^3_n\varepsilon+C^4_n+C^5_n\varepsilon^2+C^6_n\varepsilon+C^7_n+...$

$\displaystyle Daca~adunam~aceste~trei~relatii,~observam~ca~toate~combinarile, \\ \\ exceptandu-le~pe~C^1_n,~C^4_n,~C^7_n...,~vor~avea~coeficientul~ \\ \\ 1+\varepsilon+\varepsilon^2=0,~deci~se~vor~anula. \\ \\ Obtinem~P(1)+\varepsilon^2P(\varepsilon)+\varepsilon P(\varepsilon^2)=3S,~unde~S~este~suma~pe~care \\ \\ vrem~sa~o~calculam.$

$\displaystyle Deci~S=\frac{1}{3} \left( P(1)+\varepsilon^2P(\varepsilon)+\varepsilon P(\varepsilon^2) \right)= \\ \\ =\frac{1}{3} \left(2^n+\varepsilon^2(\varepsilon+1)^n+\varepsilon(\varepsilon^2+1)^n \right)= \\ \\ =\frac{1}{3} \left(2^n+\varepsilon^2 \cdot (-\varepsilon^2)^n+ \varepsilon \cdot (-\varepsilon)^n \right)= \\ \\ =\frac{1}{3} \left(2^n+(-1)^n\varepsilon^{2n+2}+(-1)^n\varepsilon^{n+1} \right)$
$\displaystyle Pentru~o~forma~finala~se~poate~inlocui~\varepsilon~cu~\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3} \\ \\ si~apoi~se~poate~utiliza~formula~lui~Moivre.$