Question-225613

Question Number 225613 by Jubr last updated on 04/Nov/25

Answered by mahdipoor last updated on 04/Nov/25

x^2 +x+1=0 ⇒((−1±(√(1^2 −4.1.1)))/(2.1))=((−1)/2)±((√3)/2)i =cos(±120)+i.sin(±120)=e^(i(±((2π)/3))) v_n =e^(iθn) +e^(i(−θ)n) =2cos(((2πn)/3)) z_n =(−1)^n v_n Σ_(0+6k) ^(5+6k) z_n =2(cos0−cos120+cos240−cos0+cos120−cos240) =0 Σ_0 ^(2023) z_n = z_n ∣_(n=2022 , 2023) =2(cos0−cos120)=3

$$\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{0}\:\Rightarrow\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}^{\mathrm{2}} −\mathrm{4}.\mathrm{1}.\mathrm{1}}}{\mathrm{2}.\mathrm{1}}=\frac{−\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$=\mathrm{cos}\left(\pm\mathrm{120}\right)+\mathrm{i}.\mathrm{sin}\left(\pm\mathrm{120}\right)=\mathrm{e}^{\mathrm{i}\left(\pm\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} \\ $$$$\mathrm{v}_{\mathrm{n}} =\mathrm{e}^{\mathrm{i}\theta\mathrm{n}} +\mathrm{e}^{\mathrm{i}\left(−\theta\right)\mathrm{n}} =\mathrm{2cos}\left(\frac{\mathrm{2}\pi\mathrm{n}}{\mathrm{3}}\right)\:\:\:\: \\ $$$$\mathrm{z}_{\mathrm{n}} =\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{v}_{\mathrm{n}} \\ $$$$\sum_{\mathrm{0}+\mathrm{6k}} ^{\mathrm{5}+\mathrm{6k}} \:\mathrm{z}_{\mathrm{n}} =\mathrm{2}\left(\mathrm{cos0}−\mathrm{cos120}+\mathrm{cos240}−\mathrm{cos0}+\mathrm{cos120}−\mathrm{cos240}\right) \\ $$$$=\mathrm{0} \\ $$$$\sum_{\mathrm{0}} ^{\mathrm{2023}} \mathrm{z}_{\mathrm{n}} =\:\mathrm{z}_{\mathrm{n}} \mid_{\mathrm{n}=\mathrm{2022}\:,\:\mathrm{2023}} =\mathrm{2}\left(\mathrm{cos0}−\mathrm{cos120}\right)=\mathrm{3} \\ $$$$ \\ $$

Commented by Frix last updated on 04/Nov/25

2Σ_0 ^n (−1)^j cos ((2πj)/3) = { ((3, n=6k+1)),((2, n=6k+2∨n=6k)),((0, n=6k+3∨n=6k+5)),((−1, n=6k+4)) :} n=2023=6×337+1 ⇒ answer is 3

$$\mathrm{2}\underset{\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{j}} \mathrm{cos}\:\frac{\mathrm{2}\pi{j}}{\mathrm{3}}\:=\begin{cases}{\mathrm{3},\:{n}=\mathrm{6}{k}+\mathrm{1}}\\{\mathrm{2},\:{n}=\mathrm{6}{k}+\mathrm{2}\vee{n}=\mathrm{6}{k}}\\{\mathrm{0},\:{n}=\mathrm{6}{k}+\mathrm{3}\vee{n}=\mathrm{6}{k}+\mathrm{5}}\\{−\mathrm{1},\:{n}=\mathrm{6}{k}+\mathrm{4}}\end{cases} \\ $$$${n}=\mathrm{2023}=\mathrm{6}×\mathrm{337}+\mathrm{1}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{3} \\ $$

Commented by Frix last updated on 05/Nov/25

��

Commented by mahdipoor last updated on 04/Nov/25