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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) =?

$${Given}\:\begin{cases}{{a}_{\mathrm{2}{n}} \:=\:{a}_{{n}} .{a}_{\mathrm{2}} \:+\mathrm{1}}\\{{a}_{\mathrm{2}{n}+\mathrm{1}} \:=\:{a}_{{n}} .{a}_{\mathrm{2}} \:−\mathrm{2}\:}\end{cases}\:{and} \\ $$ $$\:\begin{cases}{{a}_{\mathrm{7}} \:=\:\mathrm{2}}\\{\mathrm{0}<{a}_{\mathrm{1}} <\mathrm{1}}\end{cases}.\:{Find}\:{a}_{\mathrm{25}} \:=? \\ $$ $$ \\ $$

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 )) :}

$$\left(\ast\right)\:{a}_{\mathrm{25}} \:=\:{a}_{\mathrm{12}} .{a}_{\mathrm{2}} −\mathrm{2}\: \\ $$ $$\left(\ast\right)\:{a}_{\mathrm{12}} =\:{a}_{\mathrm{6}} .{a}_{\mathrm{2}} \:+\:\mathrm{1} \\ $$ $$\left(\bullet\right){a}_{\mathrm{7}} \:=\:{a}_{\mathrm{3}} .{a}_{\mathrm{2}} \:−\mathrm{2}\:\Rightarrow{a}_{\mathrm{3}} .{a}_{\mathrm{2}} =\:\mathrm{4} \\ $$ $$\rightarrow\begin{cases}{{a}_{\mathrm{3}} ={a}_{\mathrm{1}} .{a}_{\mathrm{2}} −\mathrm{2}}\\{{a}_{\mathrm{2}} \:=\:{a}_{\mathrm{1}} .{a}_{\mathrm{2}} \:+\mathrm{1}\rightarrow{a}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}−{a}_{\mathrm{1}} }}\end{cases} \\ $$ $${let}\:{a}_{\mathrm{1}} .{a}_{\mathrm{2}} \:=\:{k}\:\Rightarrow\left({k}−\mathrm{2}\right)\left({k}+\mathrm{1}\right)=\mathrm{4} \\ $$ $$\Rightarrow{k}^{\mathrm{2}} −{k}−\mathrm{6}\:=\:\mathrm{0}\:\rightarrow\begin{cases}{{k}=\mathrm{3}\rightarrow{a}_{\mathrm{2}} =\frac{\mathrm{3}}{{a}_{\mathrm{1}} }\:>\mathrm{0}}\\{{k}=−\mathrm{2}\left({reject}\right)}\end{cases} \\ $$ $$\rightarrow\begin{cases}{{a}_{\mathrm{3}} =\:\mathrm{1}}\\{{a}_{\mathrm{2}} =\:\mathrm{4}}\end{cases}\:;\:{a}_{\mathrm{6}} \:=\:{a}_{\mathrm{3}} .{a}_{\mathrm{2}} +\mathrm{1}=\:\mathrm{5} \\ $$ $$\rightarrow\begin{cases}{{a}_{\mathrm{12}} =\:{a}_{\mathrm{6}} .{a}_{\mathrm{2}} +\mathrm{1}\:=\:\mathrm{21}}\\{{a}_{\mathrm{25}} \:=\:{a}_{\mathrm{12}} .{a}_{\mathrm{2}} −\mathrm{2}=\mathrm{82}\:}\end{cases} \\ $$ $$ \\ $$

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