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Question Number 184302 by cortano1 last updated on 05/Jan/23

Answered by mr W last updated on 05/Jan/23

say f(x)=Ap^x   Ap^(x−1) +Ap^(x+1) =(√3)Ap^x   1+p^2 =(√3)p  p^2 −(√3)p+1=0  ⇒p=(((√3)±i)/2)=e^(±((πi)/6))   ⇒f(x)=Ae^((πxi)/6) +Be^(−((πxi)/6))   f(5+12r)=Ae^((π(5+12r)i)/6) +Be^(−((π(5+12r)i)/6))   f(5+12r)=Ae^((((5π)/6)+2rπ)i) +Be^(−(((5π)/6)+2rπ)i)   f(5+12r)=Ae^((((5π)/6))i) +Be^(−(((5π)/6))i) =f(5)=7  Σ_(r=0) ^(288) f(5+12r)=Σ_(r=0) ^(288) 7=289×7=2023

$${say}\:{f}\left({x}\right)={Ap}^{{x}} \\ $$$${Ap}^{{x}−\mathrm{1}} +{Ap}^{{x}+\mathrm{1}} =\sqrt{\mathrm{3}}{Ap}^{{x}} \\ $$$$\mathrm{1}+{p}^{\mathrm{2}} =\sqrt{\mathrm{3}}{p} \\ $$$${p}^{\mathrm{2}} −\sqrt{\mathrm{3}}{p}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{p}=\frac{\sqrt{\mathrm{3}}\pm{i}}{\mathrm{2}}={e}^{\pm\frac{\pi{i}}{\mathrm{6}}} \\ $$$$\Rightarrow{f}\left({x}\right)={Ae}^{\frac{\pi{xi}}{\mathrm{6}}} +{Be}^{−\frac{\pi{xi}}{\mathrm{6}}} \\ $$$${f}\left(\mathrm{5}+\mathrm{12}{r}\right)={Ae}^{\frac{\pi\left(\mathrm{5}+\mathrm{12}{r}\right){i}}{\mathrm{6}}} +{Be}^{−\frac{\pi\left(\mathrm{5}+\mathrm{12}{r}\right){i}}{\mathrm{6}}} \\ $$$${f}\left(\mathrm{5}+\mathrm{12}{r}\right)={Ae}^{\left(\frac{\mathrm{5}\pi}{\mathrm{6}}+\mathrm{2}{r}\pi\right){i}} +{Be}^{−\left(\frac{\mathrm{5}\pi}{\mathrm{6}}+\mathrm{2}{r}\pi\right){i}} \\ $$$${f}\left(\mathrm{5}+\mathrm{12}{r}\right)={Ae}^{\left(\frac{\mathrm{5}\pi}{\mathrm{6}}\right){i}} +{Be}^{−\left(\frac{\mathrm{5}\pi}{\mathrm{6}}\right){i}} ={f}\left(\mathrm{5}\right)=\mathrm{7} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{\mathrm{288}} {\sum}}{f}\left(\mathrm{5}+\mathrm{12}{r}\right)=\underset{{r}=\mathrm{0}} {\overset{\mathrm{288}} {\sum}}\mathrm{7}=\mathrm{289}×\mathrm{7}=\mathrm{2023} \\ $$

Commented by cortano1 last updated on 05/Jan/23

nice solution

$${nice}\:{solution} \\ $$

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