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Question Number 34186 by Joel578 last updated on 02/May/18

Let x_{1} = 0, x_{2} = 1 and x_{n} = \frac{1}{2} (x_{n - 1} + x_{n - 2})

Show that

x_{n} = \frac{2^{n - 1} + {(- 1)}^{n}}{3 . 2^{n - 2}}

Commented by abdo mathsup 649 cc last updated on 02/May/18

\Leftrightarrow 2 x_{n} = x_{n - 1} + x_{n - 2} \Leftrightarrow 2 x_{n + 2} = x_{n + 1} + x_{n}

\Leftrightarrow 2 x_{n + 2} - x_{n + 1} - x_{n} = 0 t h e c s r s c t e r i s t i c e q u a t i o n

i s 2 x^{2} - x - 1 = 0

Δ = 1 - 4 (2) (- 1) = 9 > 0

x_{1} = \frac{1 + 3}{4} = 1, x_{2} = \frac{1 - 3}{4} = - \frac{1}{2} t h e r o o t s a r e s i m p l e s

\Rightarrow x_{n} = α 1^{n} + β {(- \frac{1}{2})}^{n} = α + β {(- \frac{1}{2})}^{n}

x_{1} = 0 \Rightarrow α - \frac{β}{2} = 0 \Rightarrow β = 2 α

x_{2} = 1 \Rightarrow α + \frac{β}{4} = 1 \Rightarrow 4 α + β = 4 \Rightarrow 4 α + 2 α = 4 \Rightarrow

6 α = 4 \Rightarrow α = \frac{2}{3} a n d β = \frac{4}{3} \Rightarrow

x_{n} = \frac{2}{3} + \frac{4}{3} {(- \frac{1}{2})}^{n} = \frac{2}{3} + \frac{4 {(- 1)}^{n}}{3 . 2^{n}}

= \frac{1}{3} (\frac{2^{n + 1} + 4 . {(- 1)}^{n}}{2^{n}}) = \frac{4}{3} (\frac{2^{n - 1} + {(- 1)}^{n}}{2^{n}})

= \frac{2^{n - 1} + {(- 1)}^{n}}{3 . 2^{n - 2}} .

Answered by candre last updated on 02/May/18

x_{1} = \frac{2^{1 - 1} + {(- 1)}^{1}}{3 \cdot 2^{1 - 2}} = \frac{1 - 1}{3 / 2} = 0

x_{2} = \frac{2^{2 - 1} + {(- 1)}^{2}}{3 \cdot 2^{2 - 2}} = \frac{2 + 1}{3} = 1

x_{n - 1} = \frac{2^{n - 2} + {(- 1)}^{n - 1}}{3 \cdot 2^{n - 3}}

x_{n - 2} = \frac{2^{n - 3} + {(- 1)}^{n - 2}}{3 \cdot 2^{n - 4}}

\frac{x_{n - 1} + x_{n - 2}}{2} = \frac{1}{2} (\frac{2^{n - 2} + {(- 1)}^{n - 1}}{3 \cdot 2^{n - 3}} + \frac{2^{n - 3} + {(- 1)}^{n - 2}}{3 \cdot 2^{n - 4}})

= \frac{1}{2^{n + 1} \cdot 3} (\frac{2^{n - 2} - {(- 1)}^{n}}{2^{- 3}} + \frac{2^{n - 3} + {(- 1)}^{n}}{2^{- 4}})

= \frac{1}{2^{n + 1} \cdot 3} [2^{3} (2^{n - 2} - {(- 1)}^{n}) + 2^{4} (2^{n - 3} + {(- 1)}^{n})]

= \frac{2^{3} 2^{n - 2} - 2^{3} {(- 1)}^{n} + 2^{4} 2^{n - 3} + 2^{4} {(- 1)}^{n}}{2^{n + 1} \cdot 3}

= \frac{2^{n + 2} + (2^{4} - 2^{3}) {(- 1)}^{n}}{2^{n + 1} \cdot 3}

= \frac{2^{3} (2^{n - 1} + (2 - 1) {(- 1)}^{n})}{2^{n + 1} \cdot 3}

= \frac{2^{n - 1} + {(- 1)}^{n}}{2^{n - 2} \cdot 3}

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