dear-mr-w-a-n-2-a-n-1-a-n-find-a-n-

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

dear mr w

a_{n + 2} = a_{n + 1} - a_{n}

find a_{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{π}{3} \pm i \sin \frac{π}{3} = e^{\pm i \frac{π}{3}}

\Rightarrow a_{n} = {A e}^{i \frac{n π}{3}} + {B e}^{- i \frac{n π}{3}}

y o u n e e d t o k n o w a_{1} a n d a_{2} t o d e t e r m i n e

A a n d B .

Commented by jagoll last updated on 04/Apr/20

given \underset{n = 1}{\sum^{1945}} a_{n} = 2018 and

\underset{n = 1}{\sum^{2018}} a_{n} = 1945 . Evaluate ?

\underset{n = 1}{\sum^{2001}} a_{n} = ?

Commented by jagoll last updated on 04/Apr/20

how to get A & B sir ?

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

a_{n} = {A e}^{i \frac{n π}{3}} + {B e}^{- i \frac{n π}{3}} = {A p}^{n} + {B q}^{n}

\underset{n = 1}{\sum^{1945}} a_{n} = A (\underset{n = 1}{\sum^{1945}} p^{n}) + B (\underset{n = 1}{\sum^{1945}} q^{n}) = 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 (\underset{n = 1}{\sum^{2018}} p^{n}) + B (\underset{n = 1}{\sum^{2018}} q^{n}) = 1945

\Rightarrow \frac{p (p^{2018} - 1)}{p - 1} A + \frac{q (q^{2018} - 1)}{q - 1} B = 1945

\dots \dots

Commented by jagoll last updated on 04/Apr/20

waw \dots

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