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-n-if-a-0-

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} .