let u_0 =a , u_1 =b and u_(n+2) =(1/2)(u_n +u_(n+1) ) 1) find u_n interms of n 2) find lim_(n→∞) u


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Question Number 32995 by abdo imad last updated on 09/Apr/18

$l e t u_{0} = a, u_{1} = b a n d u_{n + 2} = \frac{1}{2} (u_{n} + u_{n + 1})$ $1) f i n d u_{n} i n t e r m s o f n$ $2) f i n d {l i m}_{n \to \infty} u_{n} i f a = 0$
Commented by abdo imad last updated on 10/Apr/18

$u_{n + 2} = \frac{1}{2} (u_{n} + u_{n + 1}) \Rightarrow 2 u_{n + 2} = u_{n} + u_{n + 1} \Rightarrow$ $2 u_{n + 2} - u_{n + 1} - u_{n} = 0 t h e c a r a c t e r i s t i c e q u a t i o n i s$ $2 x^{2} - x - 1 = 0 w e h a v e Δ = 1 - 4 (2) (- 1) = 9 \Rightarrow$ $x_{1} = \frac{1 + 3}{4} = 1 a n d x_{2} = \frac{1 - 3}{4} = - \frac{1}{2} \Rightarrow u_{n} = α + β {(- \frac{1}{2})}^{n}$ $u_{0} = a \Rightarrow α + β = a, u_{1} = b \Rightarrow α - \frac{β}{2} = b \Rightarrow$ $3 \frac{β}{2} = a - b \Rightarrow β = \frac{2}{3} (a - b), α = a - β = a - \frac{2}{3} (a - b)$ $= \frac{a + 2 b}{3} \Rightarrow u_{n} = \frac{a + 2 b}{3} + \frac{2 a - 2 b}{3} {(- \frac{1}{2})}^{n}$ $2) a = 0 \Rightarrow u_{n} = \frac{2 b}{3} - \frac{2 b}{3} {(- \frac{1}{2})}^{n} b u t {l i m}_{n \to \infty} {(- \frac{1}{2})}^{n} = 0 \Rightarrow$ ${l i m}_{n \to \infty} u_{n} = \frac{2 b}{3} .$
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