Question-218820

Question Number 218820 by universe last updated on 16/Apr/25

Answered by mr W last updated on 17/Apr/25

a_n =a+b−((ab)/a_(n−1) ) let a_n =(t_n /t_(n−1) ) (t_n /t_(n−1) )=a+b−((abt_(n−2) )/t_(n−1) )=(((a+b)t_(n−1) −abt_(n−2) )/t_(n−1) ) t_n =(a+b)t_(n−1) −abt_(n−2) t_n −(a+b)t_(n−1) +abt_(n−2) =0 p^2 −(a+b)p+ab=0 ⇒p_(1,2) =a,b ⇒t_n =Aa^n +Bb^n a_n =((Aa^n +Bb^n )/(Aa^(n−1) +Bb^(n−1) ))=((Ca^n +b^n )/(Ca^(n−1) +b^(n−1) )) a_1 =((Ca+b)/(C+1))=a+b ⇒C=−(a/b) ⇒a_n =((a^(n+1) −b^(n+1) )/(a^n −b^n )) ✓

$${a}_{{n}} ={a}+{b}−\frac{{ab}}{{a}_{{n}−\mathrm{1}} } \\ $$$${let}\:{a}_{{n}} =\frac{{t}_{{n}} }{{t}_{{n}−\mathrm{1}} } \\ $$$$\frac{{t}_{{n}} }{{t}_{{n}−\mathrm{1}} }={a}+{b}−\frac{{abt}_{{n}−\mathrm{2}} }{{t}_{{n}−\mathrm{1}} }=\frac{\left({a}+{b}\right){t}_{{n}−\mathrm{1}} −{abt}_{{n}−\mathrm{2}} }{{t}_{{n}−\mathrm{1}} } \\ $$$${t}_{{n}} =\left({a}+{b}\right){t}_{{n}−\mathrm{1}} −{abt}_{{n}−\mathrm{2}} \\ $$$${t}_{{n}} −\left({a}+{b}\right){t}_{{n}−\mathrm{1}} +{abt}_{{n}−\mathrm{2}} =\mathrm{0} \\ $$$${p}^{\mathrm{2}} −\left({a}+{b}\right){p}+{ab}=\mathrm{0} \\ $$$$\Rightarrow{p}_{\mathrm{1},\mathrm{2}} ={a},{b} \\ $$$$\Rightarrow{t}_{{n}} ={Aa}^{{n}} +{Bb}^{{n}} \\ $$$${a}_{{n}} =\frac{{Aa}^{{n}} +{Bb}^{{n}} }{{Aa}^{{n}−\mathrm{1}} +{Bb}^{{n}−\mathrm{1}} }=\frac{{Ca}^{{n}} +{b}^{{n}} }{{Ca}^{{n}−\mathrm{1}} +{b}^{{n}−\mathrm{1}} } \\ $$$${a}_{\mathrm{1}} =\frac{{Ca}+{b}}{{C}+\mathrm{1}}={a}+{b} \\ $$$$\Rightarrow{C}=−\frac{{a}}{{b}} \\ $$$$\Rightarrow{a}_{{n}} =\frac{{a}^{{n}+\mathrm{1}} −{b}^{{n}+\mathrm{1}} }{{a}^{{n}} −{b}^{{n}} }\:\checkmark \\ $$

Commented by mr W last updated on 17/Apr/25