Áp dụng khai triển nhị thức Newton:
[TEX](1+\sqrt3.i)^{2010}=C_{2010}^0+C_{2010}^1.\sqrt3.i+C_{2010}^2(\sqrt3i)^2+......+C_{2010}^k(\sqrt3.i)^k+...+C_{2010}^{2009}(\sqrt3.i)^{2009}+C_{2010}^{2010}(\sqrt3.i)^{2010}[/TEX] [TEX](1)[/TEX]
[TEX](1-\sqrt3.i)^{2010}=C_{2010}^0-C_{2010}^1.\sqrt3.i+C_{2010}^2(\sqrt3i)^2+..+(-1)^kC_{2010}^k(\sqrt3.i)^k+...-C_{2010}^{2009}(\sqrt3.i)^{2009}+C_{2010}^{2010}(\sqrt3.i)^{2010}[/TEX] [TEX](2)[/TEX]
[TEX](1)+(2)=C_{2010}^0+C_{2010}^2(\sqrt3i)^2+C_{2010}^4(\sqrt3i)^4+...+C_{2010}^{2k}(\sqrt3.i)^{2k}+...+C_{2010}^{2008}(\sqrt3.i)^{2008}+C_{2010}^{2010}(\sqrt3.i)^{2010}[/TEX]
[TEX]=C_{2010}^0 - 3C_{2010}^2 + 3^2C_{2010}^4+...+ (-1)^k.3^k.C_{2010}^{2k}+...+3^{1004}C_{2010}^{2008} - 3^{1005}C_{2010}^{2010}[/TEX][TEX]=2S[/TEX]
Vậy
[TEX]2S=(1+\sqrt3.i)^{2010}+(1-\sqrt3.i)^{2010}[/TEX][TEX]=(2.e^{i.\frac{\pi}{3}})^{2010}+(2.e^{i.\frac{-\pi}{3}})^{2010}[/TEX][TEX]=2^{2011}[/TEX] [TEX]S=2^{2010}[/TEX]
Chúc các bạn vui!
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
