U-n-4-n-1-1-1-4-n-U-n-1-with-U-0-1-find-U-n-in-terms-of-n-question-Q173132-reposted-

Question Number 184401 by mr W last updated on 06/Jan/23

U_{n} = (\frac{{(- 4)}^{n + 1} - 1}{1 - {(- 4)}^{n}}) U_{n - 1} w i t h U_{0} = 1

f i n d U_{n} i n t e r m s o f n

(q u e s t i o n Q 173132 r e p o s t e d)

Commented by Frix last updated on 06/Jan/23

Sir have you tried Q 178468 ?

Commented by mr W last updated on 07/Jan/23

n o t y e t s i r .

Commented by Frix last updated on 07/Jan/23

I tried but could not find any path \dots

Answered by witcher3 last updated on 06/Jan/23

\frac{U_{n}}{U_{n - 1}} = - \frac{1 - {(- 4)}^{n + 1}}{1 - {(- 4)}^{n}}, a_{n} = 1 - {(- 4)}^{n}

\frac{U_{k}}{U_{k - 1}} = - \frac{a_{k + 1}}{a_{k}}

\underset{k = 1}{\prod^{n}} \frac{U_{k}}{U_{k - 1}} = \underset{k = 1}{\prod^{n}} - \frac{a_{k + 1}}{a_{k}} \Leftrightarrow \frac{U_{n}}{U_{0}} = {(- 1)}^{n} . \frac{a_{n + 1}}{a_{1}} = \frac{{(- 1)}^{n}}{5} (1 - {(- 4)}^{n + 1})

\Leftrightarrow u_{n} = \frac{{(- 1)}^{n}}{5} (1 - {(- 4)}^{n + 1})

Commented by witcher3 last updated on 06/Jan/23

y^{'} r e W e l c o m h a v e a N i c e D a y

Commented by mr W last updated on 06/Jan/23

t h a n k s! g r e a t s o l u t i o n!

Answered by mr W last updated on 06/Jan/23

U_{n} = (\frac{{(- 4)}^{n + 1} - 1}{1 - {(- 4)}^{n}}) U_{n - 1}

\frac{U_{n}}{{(- 4)}^{n + 1} - 1} = - \frac{U_{n - 1}}{{(- 4)}^{n} - 1}

\frac{U_{n - 1}}{{(- 4)}^{n} - 1} = - \frac{U_{n - 2}}{{(- 4)}^{n - 1} - 1}

\dots \dots

\frac{U_{1}}{{(- 4)}^{2} - 1} = - \frac{U_{0}}{{(- 4)}^{1} - 1}

\frac{U_{n}}{{(- 4)}^{n + 1} - 1} = {(- 1)}^{n} \frac{U_{0}}{{(- 4)}^{1} - 1} = \frac{{(- 1)}^{n}}{- 5}

\Rightarrow U_{n} = \frac{{(- 1)}^{n} ({(- 4)}^{n + 1} - 1)}{- 5}

= \frac{4^{n + 1} + {(- 1)}^{n}}{5} ✓

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