If a_(n+1) =2−5a_n and a_4 = −8 prove that a_(43) −a


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Question Number 207021 by efronzo1 last updated on 03/May/24

$I f a_{n + 1} = 2 - 5 a_{n} a n d a_{4} = - 8$ $p r o v e t h a t a_{43} - a_{30} d i v i s i b l e b y 5$
Commented by mr W last updated on 04/May/24

$a s s h o w n i n Q 206997 w e c a n g e t$ $a_{n} = - \frac{{(- 5)}^{n - 2}}{3} + \frac{1}{3}$ $a_{43} - a_{30} = \frac{{(- 5)}^{28} - {(- 5)}^{41}}{3} = \frac{5^{28} (1 + 5^{13})}{3} \equiv$ $1 + 5^{13} = 1 + {(2 \times 3 - 1)}^{13}$ $= 1 + (m u l t i p l e o f 3) - 1$ $= (m u l t i p l e o f 3)$ $\Rightarrow a_{43} - a_{30} i s d i v i s i b l e b y 5$
	Answered by Berbere last updated on 04/May/24

	$a_{n + 1} \equiv 2 [5]; \forall n ⩾ 4 ∴ a_{n + 1} = 2 - 5 . a_{n} ∵$ $a_{43} \equiv 2 [5]; a_{30} \equiv 2 [5] \Rightarrow a_{43} - a_{30} \equiv 0 [5]$
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