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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

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

Commented by cortano1 last updated on 05/Jan/23

$$nicesolution$$

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