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[frazier]

chứng minh rằng
$C_{2014}^0 -3.C_{2014}^2 +9.C_{2014}^4- ... - 3^{1007}.C_{2014}^{2014} = -2^{2013}$

Edited by frazier, 22-05-2014 - 16:36.

[1110004]

Bạn dùng số phức dạng lượng giác là xong!

Ta có : $\left\{\begin{matrix}(1+\sqrt{3}i)^{2014}=&C_{2014}^{0}+\sqrt{3}i.C_{2014}^{1}+...+(\sqrt{3}i)^{2014}.C_{2014}^{2014} \\ (1-\sqrt{3}i)^{2014}= &C_{2014}^{0}-\sqrt{3}i.C_{2014}^{1}+...+(-\sqrt{3}i)^{2014}.C_{2014}^{2014} \end{matrix}\right.$

 

$\Rightarrow \frac{1}{2}\left ((1+\sqrt{3}i)^{2014}+ (1-\sqrt{3}i)^{2014} \right )=C_{2014}^0 -3.C_{2014}^2 +9.C_{2014}^4- ... - 3^{1007}.C_{2014}^{2014}$

 

Mà  $\frac{1}{2}\left ((1+\sqrt{3}i)^{2014}+ (1-\sqrt{3}i)^{2014} \right )= \frac{1}{2}\left ( 2^{2014}cos(\frac{2014\pi}{3})+2^{2014}cos(\frac{2014\pi}{3}) \right )=-2^{2013}$   (đpcm)

Edited by 1110004, 22-05-2014 - 16:03.

[frazier]

cám ơn nhé!