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Question Number 213802 by efronzo1 last updated on 17/Nov/24

Answered by A5T last updated on 17/Nov/24

t_1 =k−1;t_2 =k t_3 =((5k+1)/(25k−25)); t_4 =(((10k−4)/(5k−5))/(25k))=((10k−4)/(125k(k−1))) t_5 =(((−15k−4+25k^2 )/(25k(k−1)))/((5k+1)/(k−1)))=((25k^2 −15k−4)/(25k(5k+1)))=(((5k+1)(5k−4))/(25k(5k+1))) ⇒t_5 =((5k−4)/(25k));t_6 =(((10k−4)/(5k))/((10k−4)/(5k(k−1))))=k−1⇒t_7 =((5k−4)/((5k−4)/k))=k ⇒(t_1 ,t_2 ,...,t_5 )=(t_6 ,t_7 ,...,t_(10) )=...=(t_(5n−4) ,t_(5n−3) ,...,t_(5n) ) 2020=5n⇒t_(2020) =t_5 =((5(21)−4)/(25×21))=((101)/(525))

$${t}_{\mathrm{1}} ={k}−\mathrm{1};{t}_{\mathrm{2}} ={k} \\ $$$${t}_{\mathrm{3}} =\frac{\mathrm{5}{k}+\mathrm{1}}{\mathrm{25}{k}−\mathrm{25}};\:\:\:\:{t}_{\mathrm{4}} =\frac{\frac{\mathrm{10}{k}−\mathrm{4}}{\mathrm{5}{k}−\mathrm{5}}}{\mathrm{25}{k}}=\frac{\mathrm{10}{k}−\mathrm{4}}{\mathrm{125}{k}\left({k}−\mathrm{1}\right)} \\ $$$${t}_{\mathrm{5}} =\frac{\frac{−\mathrm{15}{k}−\mathrm{4}+\mathrm{25}{k}^{\mathrm{2}} }{\mathrm{25}{k}\left({k}−\mathrm{1}\right)}}{\frac{\mathrm{5}{k}+\mathrm{1}}{{k}−\mathrm{1}}}=\frac{\mathrm{25}{k}^{\mathrm{2}} −\mathrm{15}{k}−\mathrm{4}}{\mathrm{25}{k}\left(\mathrm{5}{k}+\mathrm{1}\right)}=\frac{\left(\mathrm{5}{k}+\mathrm{1}\right)\left(\mathrm{5}{k}−\mathrm{4}\right)}{\mathrm{25}{k}\left(\mathrm{5}{k}+\mathrm{1}\right)} \\ $$$$\Rightarrow{t}_{\mathrm{5}} =\frac{\mathrm{5}{k}−\mathrm{4}}{\mathrm{25}{k}};{t}_{\mathrm{6}} =\frac{\frac{\mathrm{10}{k}−\mathrm{4}}{\mathrm{5}{k}}}{\frac{\mathrm{10}{k}−\mathrm{4}}{\mathrm{5}{k}\left({k}−\mathrm{1}\right)}}={k}−\mathrm{1}\Rightarrow{t}_{\mathrm{7}} =\frac{\mathrm{5}{k}−\mathrm{4}}{\frac{\mathrm{5}{k}−\mathrm{4}}{{k}}}={k} \\ $$$$\Rightarrow\left({t}_{\mathrm{1}} ,{t}_{\mathrm{2}} ,...,{t}_{\mathrm{5}} \right)=\left({t}_{\mathrm{6}} ,{t}_{\mathrm{7}} ,...,{t}_{\mathrm{10}} \right)=...=\left({t}_{\mathrm{5}{n}−\mathrm{4}} ,{t}_{\mathrm{5}{n}−\mathrm{3}} ,...,{t}_{\mathrm{5}{n}} \right) \\ $$$$\mathrm{2020}=\mathrm{5}{n}\Rightarrow{t}_{\mathrm{2020}} ={t}_{\mathrm{5}} =\frac{\mathrm{5}\left(\mathrm{21}\right)−\mathrm{4}}{\mathrm{25}×\mathrm{21}}=\frac{\mathrm{101}}{\mathrm{525}} \\ $$

Answered by Frix last updated on 17/Nov/24

n≥1 t_(5n−4) =a t_(5n−3) =b t_(5n−2) =((5b+1)/(25a)) t_(5n−1) =((5a+5b+1)/(125ab)) t_(5n) =((5a+1)/(25b))

$${n}\geqslant\mathrm{1} \\ $$$${t}_{\mathrm{5}{n}−\mathrm{4}} ={a} \\ $$$${t}_{\mathrm{5}{n}−\mathrm{3}} ={b} \\ $$$${t}_{\mathrm{5}{n}−\mathrm{2}} =\frac{\mathrm{5}{b}+\mathrm{1}}{\mathrm{25}{a}} \\ $$$${t}_{\mathrm{5}{n}−\mathrm{1}} =\frac{\mathrm{5}{a}+\mathrm{5}{b}+\mathrm{1}}{\mathrm{125}{ab}} \\ $$$${t}_{\mathrm{5}{n}} =\frac{\mathrm{5}{a}+\mathrm{1}}{\mathrm{25}{b}} \\ $$

Answered by golsendro last updated on 17/Nov/24

t_n = ((5t_(n−1) +1)/(25t_(n−2) )) ⇒5t_n = ((5t_(n−1) +1)/(5t_(n−2) )) set a_n = 5t_n ⇒ { ((a_1 =5×20=100)),((a_2 =5×21=105)) :} a_n = ((a_(n−1) +1)/a_(n−2) ) ⇒a_3 = ((53)/(50)) ; a_4 = ((103)/(5250)) a_5 = ((101)/(105)) ; a_6 = 100 , a_7 = 105 a_n o

$$\:\:\mathrm{t}_{\mathrm{n}} =\:\frac{\mathrm{5t}_{\mathrm{n}−\mathrm{1}} +\mathrm{1}}{\mathrm{25t}_{\mathrm{n}−\mathrm{2}} }\:\Rightarrow\mathrm{5t}_{\mathrm{n}} =\:\frac{\mathrm{5t}_{\mathrm{n}−\mathrm{1}} +\mathrm{1}}{\mathrm{5t}_{\mathrm{n}−\mathrm{2}} } \\ $$$$\:\:\mathrm{set}\:{a}_{{n}} =\:\mathrm{5t}_{\mathrm{n}} \:\Rightarrow\begin{cases}{{a}_{\mathrm{1}} =\mathrm{5}×\mathrm{20}=\mathrm{100}}\\{{a}_{\mathrm{2}} =\mathrm{5}×\mathrm{21}=\mathrm{105}}\end{cases} \\ $$$$\:\:{a}_{{n}} =\:\frac{{a}_{{n}−\mathrm{1}} +\mathrm{1}}{{a}_{{n}−\mathrm{2}} }\:\Rightarrow{a}_{\mathrm{3}} =\:\frac{\mathrm{53}}{\mathrm{50}}\:;\:{a}_{\mathrm{4}} =\:\frac{\mathrm{103}}{\mathrm{5250}} \\ $$$$\:\:{a}_{\mathrm{5}} \:=\:\frac{\mathrm{101}}{\mathrm{105}}\:;\:{a}_{\mathrm{6}} =\:\mathrm{100}\:,\:{a}_{\mathrm{7}} =\:\mathrm{105} \\ $$$$\:\:\:{a}_{{n}} \:\:\cancel{\underline{\underbrace{{o}}}} \\ $$$$ \\ $$

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