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Question Number 2322 by prakash jain last updated on 15/Nov/15

a_{0} = x

a_{n + 1} = \frac{1}{1 + a_{n}}

Find possible value for x such that

a_{n} = - 1 for some n \in N .

For example :

x = - 2, a_{1} = - 1

x = \frac{- 3}{2}, a_{2} = - 1

x = ? if a_{n} = - 1

Answered by 123456 last updated on 15/Nov/15

a_{n + 1} = \frac{1}{1 + a_{n}}

a_{n + 1} + a_{n + 1} a_{n} = 1

a_{n} = \frac{1 - a_{n + 1}}{a_{n + 1}}, a_{n + 1} \neq 0

b_{n + 1} = \frac{1 - b_{n}}{b_{n}}, b_{1} = - 1

b_{2} = \frac{1 - - 1}{- 1} = \frac{2}{- 1} = - 2

b_{3} = \frac{1 - - 2}{- 2} = - \frac{3}{2}

b_{4} = \frac{1 - - \frac{3}{2}}{- \frac{3}{2}} = - \frac{5}{3}

\cdot \cdot \cdot

the sequence b_{n} gives the x you want

b_{n} \overset{?}{=} - \frac{F_{n + 1}}{F_{n}} (check it)

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