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Question Number 87398 by jagoll last updated on 04/Apr/20

$$\mathrm{dear}\mathrm{mr}\mathrm{w}$$$${\mathrm{a}}_{\mathrm{n}+2}={\mathrm{a}}_{\mathrm{n}+1}-{\mathrm{a}}_{\mathrm{n}}$$$$\mathrm{find}{\mathrm{a}}_{\mathrm{n}}$$

Commented by mr W last updated on 04/Apr/20

$$x^2-x+1=0$$$$x=\frac{1\pm i\sqrt{3}}{2}=\cos \frac{\pi }{3}\pm i\sin \frac{\pi }{3}=e^{\pm i\frac{\pi }{3}}$$$$\Rightarrow a_n={Ae}^{i\frac{n\pi }{3}}+{Be}^{-i\frac{n\pi }{3}}$$$$youneedtoknow{a}_{1}and{a}_{2}todetermine$$$$AandB.$$

Commented by jagoll last updated on 04/Apr/20

$$\mathrm{given}\underset{\mathrm{n}=1}{\sum ^{1945}}{\mathrm{a}}_{\mathrm{n}}=2018\mathrm{and}$$$$\underset{\mathrm{n}=1}{\sum ^{2018}}{\mathrm{a}}_{\mathrm{n}}=1945.\mathrm{Evaluate}?$$$$\underset{\mathrm{n}=1}{\sum ^{2001}}{\mathrm{a}}_{\mathrm{n}}=?$$

Commented by jagoll last updated on 04/Apr/20

$$\mathrm{how}\mathrm{to}\mathrm{get}\mathrm{A}\mathrm{\&}\mathrm{B}\mathrm{sir}?$$

Commented by mr W last updated on 04/Apr/20

$$a_n={Ae}^{i\frac{n\pi }{3}}+{Be}^{-i\frac{n\pi }{3}}={Ap}^n+{Bq}^n$$$$\underset{n=1}{\sum ^{1945}}{a}_n=A\left(\underset{n=1}{\sum ^{1945}}{p}^n\right)+B\left(\underset{n=1}{\sum ^{1945}}{q}^n\right)=2018$$$$\Rightarrow \frac{p(p^{1945}-1)}{p-1}A+\frac{q(q^{1945}-1)}{q-1}B=2018$$$$\underset{n=1}{\sum ^{2018}}{a}_n=A\left(\underset{n=1}{\sum ^{2018}}{p}^n\right)+B\left(\underset{n=1}{\sum ^{2018}}{q}^n\right)=1945$$$$\Rightarrow \frac{p(p^{2018}-1)}{p-1}A+\frac{q(q^{2018}-1)}{q-1}B=1945$$$$......$$

Commented by jagoll last updated on 04/Apr/20

$$\mathrm{waw}...$$

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