Question

Sa se demonstreze egalitatea. Subpunctul a) - clasa a 10-a, Binomul lui newton

Answer 1

$a)\quad C_{n}^0+C_{n}^3+C_{n}^6 +... = ?\\ \\ \omega^2+\omega +1 = 0\Big|\cdot (\omega -1) \Rightarrow \omega ^3 = 1\\ \\ \text{Avem: }\\ \omega =-\dfrac{1}{2}\pm \dfrac{\sqrt 3}{2}i,~~\omega^3 = 1\\ \\ (1+x)^n = C_{n}^0+C_{n}^1 x+C_{n}^2 x+C_{n}^3 x+...\\ \\ \text{Punem } x = 1, x = \omega \text{ si }x = \omega^2 \text{ in identitatea de mai sus:}$

$2^n = C_n^0+C_n^1+C_n^2+C_n^3+C_n^4+...~~~(i)\\\\(1+\omega)^n=C_n^0+C_n^1\omega+C_n^2\omega^2+C_{n}^3\omega^3+...~~~(ii) \\ \\ (1+\omega^2)^n=C_n^0+C_n^1\omega^2+C_{n}^2\omega+C_{n}^3 +...~~~(iii)$

$\text{Adunam (i), (ii) si (iii) si obtinem:} \\ \\ 3C_n^0+C_n^1(1+\omega+\omega^2)+C_n^2(1+\omega^2+\omega)+3C_{n}^3+...= \\ =2^n+(1+\omega)^n+(1+\omega^2)^n\\ \\ 1+\omega+\omega^2 = 0\\ \\ 1+\omega= \dfrac{1}{2}\pm \dfrac{\sqrt 3}{2},\quad 1+\omega^2 =-\omega = \dfrac{1}{2}\mp \dfrac{\sqrt 3}{2}i$

$\Rightarrow 3C_n^0+3C_n^3+3C_n^6+... = 2^n+\Big(\dfrac{1}{2}\pm \dfrac{\sqrt 3}{2}i\Big)^n+\Big(\dfrac{1}{2}\mp \dfrac{\sqrt 3}{2}i\Big)^n\\ \\ \Rightarrow 3(C_n^0+C_n^3+C_n^6+...) = 2^n+(\cos\frac{\pi}{3}\pm i\sin\frac{\pi}{3})^n+(\cos\frac{\pi}{3}\mp i\sin\frac{\pi}{3})^n\\ \\ 3(C_n^0+C_n^3+C_n^6+...) = 2^n +\cos\frac{n\pi}{3}\pm i\sin \frac{n\pi}{3}+\cos\frac{n\pi}{3}\mp i\sin \frac{n\pi}{3} \\ \\ 3(C_n^0+C_n^3+C_n^6+...) = 2^n+2\cos\frac{n\pi}{3}$

$\Rightarrow \boxed{C_n^0+C_n^3+C_n^6+... = \dfrac{1}{3}\left(2^n+2\cos\dfrac{n\pi}{3}\right)}$