Given { ((a_(2n) = a_n .a_2 +1)),((a_(2n+1) = a_n .a_2 −2 )) :} and { ((a_7 = 2)),((0<a_1 <1)) :}. Find a


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Question Number 137584 by bramlexs22 last updated on 04/Apr/21
$Given { ((a_(2n) = a_n .a_2 +1)),((a_(2n+1) = a_n .a_2 −2 )) :} and { ((a_7 = 2)),((0<a_1 <1)) :}. Find a_(25) =?$
$G i v e n {\begin{cases} a_{2 n} = a_{n} . a_{2} + 1 \\ a_{2 n + 1} = a_{n} . a_{2} - 2 \end{cases} a n d$ ${\begin{cases} a_{7} = 2 \\ 0 < a_{1} < 1 \end{cases} . F i n d a_{25} = ?$
	Answered by bemath last updated on 04/Apr/21
	$(∗) a_(25) = a_(12) .a_2 −2 (∗) a_(12) = a_6 .a_2 + 1 (•)a_7 = a_3 .a_2 −2 ⇒a_3 .a_2 = 4 → { ((a_3 =a_1 .a_2 −2)),((a_2 = a_1 .a_2 +1→a_2 =(1/(1−a_1 )))) :} let a_1 .a_2 = k ⇒(k−2)(k+1)=4 ⇒k^2 −k−6 = 0 → { ((k=3→a_2 =(3/a_1 ) >0)),((k=−2(reject))) :} → { ((a_3 = 1)),((a_2 = 4)) :} ; a_6 = a_3 .a_2 +1= 5 → { ((a_(12) = a_6 .a_2 +1 = 21)),((a_(25) = a_(12) .a_2 −2=82 )) :}$
	$() a_{25} = a_{12} . a_{2} - 2$ $() a_{12} = a_{6} . a_{2} + 1$ $(∙) a_{7} = a_{3} . a_{2} - 2 \Rightarrow a_{3} . a_{2} = 4$ $\to {\begin{cases} a_{3} = a_{1} . a_{2} - 2 \\ a_{2} = a_{1} . a_{2} + 1 \to a_{2} = \frac{1}{1 - a_{1}} \end{cases}$ $l e t a_{1} . a_{2} = k \Rightarrow (k - 2) (k + 1) = 4$ $\Rightarrow k^{2} - k - 6 = 0 \to {\begin{cases} k = 3 \to a_{2} = \frac{3}{a_{1}} > 0 \\ k = - 2 (r e j e c t) \end{cases}$ $\to {\begin{cases} a_{3} = 1 \\ a_{2} = 4 \end{cases}; a_{6} = a_{3} . a_{2} + 1 = 5$ $\to {\begin{cases} a_{12} = a_{6} . a_{2} + 1 = 21 \\ a_{25} = a_{12} . a_{2} - 2 = 82 \end{cases}$
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