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Question Number 207423 by efronzo1 last updated on 14/May/24

$⇃ \underset{―}{}$
	Answered by mr W last updated on 14/May/24

	$l e t a_{n} = b_{n} + k$ $b_{n + 1} + k = - \frac{2}{3} + 2 b_{n} + 2 k$ $b_{n + 1} = 2 b_{n} + k - \frac{2}{3}$ $s e t k - \frac{2}{3} = 0, i . e . k = \frac{2}{3}$ $b_{n} = 2 b_{n - 1} = 2^{2} b_{n - 2} = . . . = 2^{n - 3} b_{3}$ $= 2^{n - 3} (a_{3} - k) = 2^{n - 3} (2 - \frac{2}{3}) = \frac{2^{n - 1}}{3}$ $\Rightarrow a_{n} = b_{n} + k = \frac{2^{n - 1}}{3} + \frac{2}{3}$ $. . . .$
	Answered by AliJumaa last updated on 15/May/24
	$i have been derrived this form : { ((u_(n+1) =au_n +b)),((u_m =k)) :} ⇒ u_n =a^n (((k+(b/(a−1)))/a^m ))−(b/(a−1)) a_n =2^n (((2−(2/3))/8))+(2/3)=2^n ((1/6))+(2/3) 1:a_5 =2^5 ((1/6))+(2/3)=6 FALS 2: a_n ≩0 ∀n∈N TRUE i will complete next time$
	$i h a v e b e e n d e r r i v e d t h i s f o r m :$ ${\begin{cases} u_{n + 1} = {a u}_{n} + b \\ u_{m} = k \end{cases}$ $\Rightarrow u_{n} = a^{n} (\frac{k + \frac{b}{a - 1}}{a^{m}}) - \frac{b}{a - 1}$ $a_{n} = 2^{n} (\frac{2 - \frac{2}{3}}{8}) + \frac{2}{3} = 2^{n} (\frac{1}{6}) + \frac{2}{3}$ $1 : a_{5} = 2^{5} (\frac{1}{6}) + \frac{2}{3} = 6 FALS$ $2 : a_{n} ≩ 0 \forall n \in N TRUE$ $i w i l l c o m p l e t e n e x t t i m e$
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