let-give-the-sequence-u-n-u-0-1-and-u-1-2-and-n-N-2u-n-2-3-u-n-1-u-n-let-give-the-sequence-v-n-v-n-u-n-1-u-n-1-prove-that-v-n-is-geometric-find-v-n-in-terms-of-n-2

Question Number 29148 by abdo imad last updated on 04/Feb/18

l e t g i v e t h e s e q u e n c e (u_{n}) / u_{0} = 1 a n d u_{1} = 2 a n d

\forall n \in N 2 u_{n + 2} = 3 u_{n + 1} - u_{n} . l e t g i v e t h e s e q u e n c e (v_{n}) /

v_{n} = u_{n + 1} - u_{n} .

1) p r o v e t h a t (v_{n}) i s g e o m e t r i c . f i n d v_{n} i n t e r m s o f n

2) f i n d u_{n} i n t e r m s o f n .

Commented by abdo imad last updated on 08/Feb/18

1) v_{n + 1} = u_{n + 2} - u_{n + 1} = \frac{3}{2} u_{n + 1} - \frac{1}{2} u_{n} - u_{n + 1}

= \frac{1}{2} (u_{n + 1} - u_{n}) = \frac{1}{2} v_{n} s o (v_{n}) i s a g e o m e t r i c p r o g r e s s i o n

w i t h r a i z o n q = \frac{1}{2} \Rightarrow v_{n} = v_{0} q^{n} b u t v_{0} = u_{1} - u_{0} = 1

\Rightarrow v_{n} = {(\frac{1}{2})}^{n}

2) \sum_{k = 0}^{n - 1} v_{k} = \sum_{k = 0}^{n - 1} (u_{k + 1} - u_{k}) = u_{1} - u_{0} + u_{2} - u_{1} + \dots + u_{n} - u_{n - 1}

= u_{n} - u_{0} \Rightarrow u_{n} = 1 + \sum_{k = 0}^{n - 1} {(\frac{1}{2})}^{k} = 1 + \frac{1 - {(\frac{1}{2})}^{n}}{1 - \frac{1}{2}}

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

u_{n} = 3 - \frac{1}{2^{n - 1}} .