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Question-159488

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Question Number 159488 by Tawa11 last updated on 17/Nov/21
Answered by mr W last updated on 17/Nov/21
T_n =4T_(n−1) −3 let T_n =S_n +A S_n +A=4(S_(n−1) +A)−3 S_n =4S_(n−1) +(3A−3) let 3A−3=0, i.e. A=1 S_n =4S_(n−1) this is a G.P. ⇒S_n =4^(n−2) S_2 ⇒(T_n −1)=4^(n−2) (T_2 −1) ⇒T_n =4^(n−2) (T_2 −1)+1 ⇒T_3 =4(T_2 −1)+1=4T_2 −3 2T_3 =5T_2 as given 2(4T_2 −3)=5T_2 ⇒T_2 =2 ⇒T_n =4^(n−2) (2−1)+1 ⇒T_n =4^(n−2) +1 T_1 =4^(−1) +1=(5/4) T_2 =4^0 +1=2 T_3 =4+1=5 ......
$${T}_{{n}} =\mathrm{4}{T}_{{n}−\mathrm{1}} −\mathrm{3} \\ $$$${let}\:{T}_{{n}} ={S}_{{n}} +{A} \\ $$$${S}_{{n}} +{A}=\mathrm{4}\left({S}_{{n}−\mathrm{1}} +{A}\right)−\mathrm{3} \\ $$$${S}_{{n}} =\mathrm{4}{S}_{{n}−\mathrm{1}} +\left(\mathrm{3}{A}−\mathrm{3}\right) \\ $$$${let}\:\mathrm{3}{A}−\mathrm{3}=\mathrm{0},\:{i}.{e}.\:{A}=\mathrm{1} \\ $$$${S}_{{n}} =\mathrm{4}{S}_{{n}−\mathrm{1}} \\ $$$${this}\:{is}\:{a}\:{G}.{P}. \\ $$$$\Rightarrow{S}_{{n}} =\mathrm{4}^{{n}−\mathrm{2}} {S}_{\mathrm{2}} \\ $$$$\Rightarrow\left({T}_{{n}} −\mathrm{1}\right)=\mathrm{4}^{{n}−\mathrm{2}} \left({T}_{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\Rightarrow{T}_{{n}} =\mathrm{4}^{{n}−\mathrm{2}} \left({T}_{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1} \\ $$$$\Rightarrow{T}_{\mathrm{3}} =\mathrm{4}\left({T}_{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}=\mathrm{4}{T}_{\mathrm{2}} −\mathrm{3} \\ $$$$\mathrm{2}{T}_{\mathrm{3}} =\mathrm{5}{T}_{\mathrm{2}} \:{as}\:{given} \\ $$$$\mathrm{2}\left(\mathrm{4}{T}_{\mathrm{2}} −\mathrm{3}\right)=\mathrm{5}{T}_{\mathrm{2}} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow{T}_{{n}} =\mathrm{4}^{{n}−\mathrm{2}} \left(\mathrm{2}−\mathrm{1}\right)+\mathrm{1} \\ $$$$\Rightarrow{T}_{{n}} =\mathrm{4}^{{n}−\mathrm{2}} +\mathrm{1} \\ $$$${T}_{\mathrm{1}} =\mathrm{4}^{−\mathrm{1}} +\mathrm{1}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${T}_{\mathrm{2}} =\mathrm{4}^{\mathrm{0}} +\mathrm{1}=\mathrm{2} \\ $$$${T}_{\mathrm{3}} =\mathrm{4}+\mathrm{1}=\mathrm{5} \\ $$$$…… \\ $$
Commented by Tawa11 last updated on 18/Nov/21
God bless you sir. I really appreciate ...
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:… \\ $$
Answered by Rasheed.Sindhi last updated on 18/Nov/21
t_n =4t_(n−1) −3 t_3 =4t_2 −3 ∧ 2t_3 =5t_2 ⇒2(4t_2 −3)=5t_2 8t_2 −5t_2 =6 t_2 =2 t_3 =4(2)−3=5 t_2 =4t_1 −3=2 t_1 =(5/4) t_1 ,t_2 ,t_3 =(5/4),2,5
$$\mathrm{t}_{\mathrm{n}} =\mathrm{4t}_{\mathrm{n}−\mathrm{1}} −\mathrm{3} \\ $$$$\mathrm{t}_{\mathrm{3}} =\mathrm{4t}_{\mathrm{2}} −\mathrm{3}\:\wedge\:\mathrm{2t}_{\mathrm{3}} =\mathrm{5t}_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\left(\mathrm{4t}_{\mathrm{2}} −\mathrm{3}\right)=\mathrm{5t}_{\mathrm{2}} \\ $$$$\mathrm{8t}_{\mathrm{2}} −\mathrm{5t}_{\mathrm{2}} =\mathrm{6} \\ $$$$\mathrm{t}_{\mathrm{2}} =\mathrm{2} \\ $$$$\mathrm{t}_{\mathrm{3}} =\mathrm{4}\left(\mathrm{2}\right)−\mathrm{3}=\mathrm{5} \\ $$$$\mathrm{t}_{\mathrm{2}} =\mathrm{4t}_{\mathrm{1}} −\mathrm{3}=\mathrm{2} \\ $$$$\mathrm{t}_{\mathrm{1}} =\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{t}_{\mathrm{1}} ,\mathrm{t}_{\mathrm{2}} ,\mathrm{t}_{\mathrm{3}} =\frac{\mathrm{5}}{\mathrm{4}},\mathrm{2},\mathrm{5} \\ $$
Commented by Tawa11 last updated on 18/Nov/21
God bless you sir, I really appreciate.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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