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If-T-n-1-1-1-2-T-n-Find-a-formular-for-T-n-in-terms-of-n-and-find-the-sum-of-first-n-terms-

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Question Number 75327 by TawaTawa last updated on 09/Dec/19
If T_(n + 1) = 1 + (1/2)T_n Find a formular for T_n in terms of n and find the sum of first n terms
$$\mathrm{If}\:\:\:\:\:\mathrm{T}_{\mathrm{n}\:+\:\mathrm{1}} \:\:\:=\:\:\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{T}_{\mathrm{n}} \\ $$$$\mathrm{Find}\:\mathrm{a}\:\mathrm{formular}\:\mathrm{for}\:\:\mathrm{T}_{\mathrm{n}} \:\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\:\mathrm{n} \\ $$$$\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{n}\:\mathrm{terms} \\ $$
Commented by malwaan last updated on 11/Dec/19
T_1 = 4 T_2 = 1 +(1/2)×4= 1+2 = 3 T_3 = 1+(1/2)×3 = 1 + (3/2) =(5/2) T_4 = 1 +(1/2)×(5/2)=1+(5/4)=(9/4) T_5 = 1+(9/8)=((17)/8)=((2^4 +1)/2^3 )=((2^(5−1) +1)/2^(5−2) ) ⇒T_n = ((2^(n−1) +1)/2^(n−2) )
$$\boldsymbol{{T}}_{\mathrm{1}} \:=\:\mathrm{4} \\ $$$$\boldsymbol{{T}}_{\mathrm{2}} \:=\:\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}=\:\mathrm{1}+\mathrm{2}\:=\:\mathrm{3} \\ $$$$\boldsymbol{{T}}_{\mathrm{3}} \:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{3}\:=\:\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\boldsymbol{{T}}_{\mathrm{4}} \:=\:\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{5}}{\mathrm{2}}=\mathrm{1}+\frac{\mathrm{5}}{\mathrm{4}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\boldsymbol{{T}}_{\mathrm{5}} \:=\:\mathrm{1}+\frac{\mathrm{9}}{\mathrm{8}}=\frac{\mathrm{17}}{\mathrm{8}}=\frac{\mathrm{2}^{\mathrm{4}} +\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }=\frac{\mathrm{2}^{\mathrm{5}−\mathrm{1}} +\mathrm{1}}{\mathrm{2}^{\mathrm{5}−\mathrm{2}} } \\ $$$$\Rightarrow\boldsymbol{{T}}_{\boldsymbol{{n}}} \:=\:\frac{\mathrm{2}^{\boldsymbol{{n}}−\mathrm{1}} +\mathrm{1}}{\mathrm{2}^{\boldsymbol{{n}}−\mathrm{2}} } \\ $$
Commented by mr W last updated on 10/Dec/19
please check the question: when giving a recursive sequence, the first term T_1 must be given in order to determine all terms.
$${please}\:{check}\:{the}\:{question}: \\ $$$${when}\:{giving}\:{a}\:{recursive}\:{sequence}, \\ $$$${the}\:{first}\:{term}\:{T}_{\mathrm{1}} \:{must}\:{be}\:{given}\:{in} \\ $$$${order}\:{to}\:{determine}\:{all}\:{terms}. \\ $$
Commented by TawaTawa last updated on 10/Dec/19
Ohh, yes sir, T_1 = 4 Please help me edit it in your work. Thanks sir
$$\mathrm{Ohh},\:\:\mathrm{yes}\:\mathrm{sir},\:\:\:\:\mathrm{T}_{\mathrm{1}} \:\:=\:\:\mathrm{4} \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{edit}\:\mathrm{it}\:\mathrm{in}\:\mathrm{your}\:\mathrm{work}.\:\mathrm{Thanks}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 10/Dec/19
say T_n =Ap^n +B T_(n+1) =Ap^(n+1) +B Ap^(n+1) +B=1+((Ap^n )/2)+(B/2) ⇒Ap^n (p−(1/2))+(B/2)=1 ⇒p−(1/2)=0 ⇒p=(1/2) ⇒(B/2)=1 ⇒B=2 ⇒T_n =(A/2^n )+2 if T_1 =4, 4=(A/2)+2 ⇒A=4 ⇒T_n =2+(1/2^(n−2) ) Σ_(k=1) ^n T_k =2n+((2(1−(1/2^n )))/(1−(1/2)))=2(n+2)−(1/2^(n−2) )
$${say}\:{T}_{{n}} ={Ap}^{{n}} +{B} \\ $$$${T}_{{n}+\mathrm{1}} ={Ap}^{{n}+\mathrm{1}} +{B} \\ $$$${Ap}^{{n}+\mathrm{1}} +{B}=\mathrm{1}+\frac{{Ap}^{{n}} }{\mathrm{2}}+\frac{{B}}{\mathrm{2}} \\ $$$$\Rightarrow{Ap}^{{n}} \left({p}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{{B}}{\mathrm{2}}=\mathrm{1} \\ $$$$\Rightarrow{p}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}\:\Rightarrow{p}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{B}}{\mathrm{2}}=\mathrm{1}\:\Rightarrow{B}=\mathrm{2} \\ $$$$\Rightarrow\boldsymbol{{T}}_{\boldsymbol{{n}}} =\frac{\boldsymbol{{A}}}{\mathrm{2}^{\boldsymbol{{n}}} }+\mathrm{2} \\ $$$${if}\:{T}_{\mathrm{1}} =\mathrm{4}, \\ $$$$\mathrm{4}=\frac{{A}}{\mathrm{2}}+\mathrm{2} \\ $$$$\Rightarrow{A}=\mathrm{4} \\ $$$$\Rightarrow\boldsymbol{{T}}_{\boldsymbol{{n}}} =\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}^{\boldsymbol{{n}}−\mathrm{2}} } \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{T}_{{k}} =\mathrm{2}{n}+\frac{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{2}\left({n}+\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{2}} } \\ $$
Commented by TawaTawa last updated on 10/Dec/19
God bless you sir. I appreciate your time
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time} \\ $$
Commented by TawaTawa last updated on 10/Dec/19
I really appreciate sir. God bless you more.
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{more}. \\ $$
Commented by TawaTawa last updated on 10/Dec/19
Sir, is there any hint on how we assume T_n = Ap^n + B. Or it is general for such questions
$$\mathrm{Sir},\:\mathrm{is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{hint}\:\mathrm{on}\:\mathrm{how}\:\mathrm{we}\:\mathrm{assume}\:\:\:\mathrm{T}_{\mathrm{n}} \:\:=\:\:\mathrm{Ap}^{\mathrm{n}} \:+\:\mathrm{B}. \\ $$$$\mathrm{Or}\:\mathrm{it}\:\mathrm{is}\:\mathrm{general}\:\mathrm{for}\:\mathrm{such}\:\mathrm{questions} \\ $$
Commented by TawaTawa last updated on 10/Dec/19
And please sir, why is p − (1/2) = 0 and (B/2) = 1
$$\mathrm{And}\:\mathrm{please}\:\mathrm{sir},\:\:\mathrm{why}\:\mathrm{is}\:\:\:\mathrm{p}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\:=\:\:\mathrm{0}\:\:\:\:\:\mathrm{and}\:\:\:\frac{\mathrm{B}}{\mathrm{2}}\:\:=\:\:\mathrm{1} \\ $$
Commented by mind is power last updated on 10/Dec/19
T_(n+1) =aT_n +b if a=1⇔T_(n+1) −T_n =b⇒Σ_(k=1) ^(n−1) (T_(k+1) −T_k )=Σ_(k=1) ^(n−1) b ⇒T_n =b(n−1)+T_1 if a≠1 let x=ax+b ⇒x=(b/(1−a)) let T_n =W_n +(b/(1−a)) ⇒W_(n+1) +(b/(1−a))=a(W_n +(b/(1−a)))+b ⇒W_(n+1) +((ab)/(1−a))=aW_n +((ab)/(1−a))+b=aW_n +(b/(1−a)) ⇒W_(n+1) =aW_n ⇒W_n =W_1 (a)^(n−1) T_n =(b/(1−a))+W_1 (a^(n−1) ) W_1 =T_1 −(b/(1−a))
$$\mathrm{T}_{\mathrm{n}+\mathrm{1}} =\mathrm{aT}_{\mathrm{n}} +\mathrm{b} \\ $$$$\mathrm{if}\:\mathrm{a}=\mathrm{1}\Leftrightarrow\mathrm{T}_{\mathrm{n}+\mathrm{1}} −\mathrm{T}_{\mathrm{n}} =\mathrm{b}\Rightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\left(\mathrm{T}_{\mathrm{k}+\mathrm{1}} −\mathrm{T}_{\mathrm{k}} \right)=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{b} \\ $$$$\Rightarrow\mathrm{T}_{\mathrm{n}} =\mathrm{b}\left(\mathrm{n}−\mathrm{1}\right)+\mathrm{T}_{\mathrm{1}} \\ $$$$\mathrm{if}\:\mathrm{a}\neq\mathrm{1} \\ $$$$\mathrm{let}\:\mathrm{x}=\mathrm{ax}+\mathrm{b}\:\Rightarrow\mathrm{x}=\frac{\mathrm{b}}{\mathrm{1}−\mathrm{a}} \\ $$$$\mathrm{let}\: \\ $$$$\mathrm{T}_{\mathrm{n}} =\mathrm{W}_{\mathrm{n}} +\frac{\mathrm{b}}{\mathrm{1}−\mathrm{a}} \\ $$$$\Rightarrow\mathrm{W}_{\mathrm{n}+\mathrm{1}} +\frac{\mathrm{b}}{\mathrm{1}−\mathrm{a}}=\mathrm{a}\left(\mathrm{W}_{\mathrm{n}} +\frac{\mathrm{b}}{\mathrm{1}−\mathrm{a}}\right)+\mathrm{b} \\ $$$$\Rightarrow\mathrm{W}_{\mathrm{n}+\mathrm{1}} +\frac{\mathrm{ab}}{\mathrm{1}−\mathrm{a}}=\mathrm{aW}_{\mathrm{n}} +\frac{\mathrm{ab}}{\mathrm{1}−\mathrm{a}}+\mathrm{b}=\mathrm{aW}_{\mathrm{n}} +\frac{\mathrm{b}}{\mathrm{1}−\mathrm{a}} \\ $$$$\Rightarrow\mathrm{W}_{\mathrm{n}+\mathrm{1}} =\mathrm{aW}_{\mathrm{n}} \Rightarrow\mathrm{W}_{\mathrm{n}} =\mathrm{W}_{\mathrm{1}} \left(\mathrm{a}\right)^{\mathrm{n}−\mathrm{1}} \\ $$$$\mathrm{T}_{\mathrm{n}} =\frac{\mathrm{b}}{\mathrm{1}−\mathrm{a}}+\mathrm{W}_{\mathrm{1}} \left(\mathrm{a}^{\mathrm{n}−\mathrm{1}} \right) \\ $$$$\mathrm{W}_{\mathrm{1}} =\mathrm{T}_{\mathrm{1}} −\frac{\mathrm{b}}{\mathrm{1}−\mathrm{a}} \\ $$$$ \\ $$
Commented by TawaTawa last updated on 10/Dec/19
God bless you sir, thanks for your time
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time} \\ $$

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