Menu Close

Find-the-value-of-t-t-1-3-2-9-3-27-n-3-n-

Tinku Tara 0 Comments
Pin




Question Number 198931 by necx122 last updated on 25/Oct/23
Find the value of t: t = (1/3)+(2/9)+(3/(27))+.......+(n/3^n )+.....
$${Find}\:{the}\:{value}\:{of}\:{t}:\: \\ $$$${t}\:=\:\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{9}}+\frac{\mathrm{3}}{\mathrm{27}}+…….+\frac{{n}}{\mathrm{3}^{{n}} }+….. \\ $$
Answered by cortano12 last updated on 26/Oct/23
t=(1/3) +(2/9) +(3/(27)) +…+ (n/3^n ) let u=(1/3) t = u+2u^2 +3u^3 +…+ n u^n t = u(1+2u+3u^2 +…+n u^(n−1) ) v=1+2u+3u^2 +…+n u^(n−1) ∫ v du = u+u^2 +u^3 +…+u^n ∫ v du = ((u(u^n −1))/(u−1))=((u^(n+1) −u)/(u−1)) v = (([(n+1)u^n −1](u−1)−(u^(n+1) −u))/((u−1)^2 )) t= u [(((n+1)u^(n+1) −(n+1)u^n −u+1−u^(n+1) +u)/((u−1)^2 )) ] t= u[ ((n u^(n+1) −(n+1)u^n +1 )/((u−1)^2 )) ]; n=(1/3) t = 3 [ ((n((1/3))^(n+1) −(n+1)((1/3))^n +1)/4) ] t =((n.3^(−n) −(n+1)3^(1−n) +3)/4) t = ((n−3(n+1)+3^(n+1) )/(4.3^n ))=((3^(n+1) −2n−3)/(4.3^n ))
$$\:\mathrm{t}=\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{2}}{\mathrm{9}}\:+\frac{\mathrm{3}}{\mathrm{27}}\:+\ldots+\:\frac{\mathrm{n}}{\mathrm{3}^{\mathrm{n}} }\: \\ $$$$\:\mathrm{let}\:\mathrm{u}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\mathrm{t}\:=\:\mathrm{u}+\mathrm{2u}^{\mathrm{2}} +\mathrm{3u}^{\mathrm{3}} \:+\ldots+\:\mathrm{n}\:\mathrm{u}^{\mathrm{n}} \: \\ $$$$\:\:\mathrm{t}\:=\:\mathrm{u}\left(\mathrm{1}+\mathrm{2u}+\mathrm{3u}^{\mathrm{2}} +\ldots+\mathrm{n}\:\mathrm{u}^{\mathrm{n}−\mathrm{1}} \right) \\ $$$$\:\:\mathrm{v}=\mathrm{1}+\mathrm{2u}+\mathrm{3u}^{\mathrm{2}} +\ldots+\mathrm{n}\:\mathrm{u}^{\mathrm{n}−\mathrm{1}} \\ $$$$\:\:\int\:\mathrm{v}\:\mathrm{du}\:=\:\mathrm{u}+\mathrm{u}^{\mathrm{2}} +\mathrm{u}^{\mathrm{3}} +\ldots+\mathrm{u}^{\mathrm{n}} \\ $$$$\:\:\:\int\:\mathrm{v}\:\mathrm{du}\:=\:\frac{\mathrm{u}\left(\mathrm{u}^{\mathrm{n}} −\mathrm{1}\right)}{\mathrm{u}−\mathrm{1}}=\frac{\mathrm{u}^{\mathrm{n}+\mathrm{1}} −\mathrm{u}}{\mathrm{u}−\mathrm{1}} \\ $$$$\:\:\:\mathrm{v}\:=\:\frac{\left[\left(\mathrm{n}+\mathrm{1}\right)\mathrm{u}^{\mathrm{n}} −\mathrm{1}\right]\left(\mathrm{u}−\mathrm{1}\right)−\left(\mathrm{u}^{\mathrm{n}+\mathrm{1}} −\mathrm{u}\right)}{\left(\mathrm{u}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\mathrm{t}=\:\mathrm{u}\:\left[\frac{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{u}^{\mathrm{n}+\mathrm{1}} −\left(\mathrm{n}+\mathrm{1}\right)\mathrm{u}^{\mathrm{n}} −\mathrm{u}+\mathrm{1}−\mathrm{u}^{\mathrm{n}+\mathrm{1}} +\mathrm{u}}{\left(\mathrm{u}−\mathrm{1}\right)^{\mathrm{2}} }\:\right] \\ $$$$\:\:\mathrm{t}=\:\mathrm{u}\left[\:\frac{\mathrm{n}\:\mathrm{u}^{\mathrm{n}+\mathrm{1}} −\left(\mathrm{n}+\mathrm{1}\right)\mathrm{u}^{\mathrm{n}} +\mathrm{1}\:}{\left(\mathrm{u}−\mathrm{1}\right)^{\mathrm{2}} }\:\right];\:\mathrm{n}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\mathrm{t}\:=\:\mathrm{3}\:\left[\:\frac{\mathrm{n}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{n}+\mathrm{1}} −\left(\mathrm{n}+\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{n}} +\mathrm{1}}{\mathrm{4}}\:\right] \\ $$$$\:\:\mathrm{t}\:=\frac{\mathrm{n}.\mathrm{3}^{−\mathrm{n}} −\left(\mathrm{n}+\mathrm{1}\right)\mathrm{3}^{\mathrm{1}−\mathrm{n}} +\mathrm{3}}{\mathrm{4}} \\ $$$$\:\:\mathrm{t}\:=\:\frac{\mathrm{n}−\mathrm{3}\left(\mathrm{n}+\mathrm{1}\right)+\mathrm{3}^{\mathrm{n}+\mathrm{1}} }{\mathrm{4}.\mathrm{3}^{\mathrm{n}} }=\frac{\mathrm{3}^{\mathrm{n}+\mathrm{1}} −\mathrm{2n}−\mathrm{3}}{\mathrm{4}.\mathrm{3}^{\mathrm{n}} } \\ $$$$ \\ $$
Commented by necx122 last updated on 26/Oct/23
Thank you for this solution. I'll look into it.
Answered by Rasheed.Sindhi last updated on 26/Oct/23
t = (1/3)+(2/9)+(3/(27))+.......+(n/3^n )+..... (1/3)t=(1/9)+(2/(27))+(3/(81))+...+(n/3^(n+1) )+... t−(1/3)t=(1/3)+(1/9)+(1/(27))+.... (2/3)t=((1/3)/(1−1/3))=(1/3)×(3/2)=(1/2) t=(1/2)×(3/2)=(3/4)
$${t}\:=\:\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{9}}+\frac{\mathrm{3}}{\mathrm{27}}+…….+\frac{{n}}{\mathrm{3}^{{n}} }+….. \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}{t}=\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{2}}{\mathrm{27}}+\frac{\mathrm{3}}{\mathrm{81}}+…+\frac{{n}}{\mathrm{3}^{{n}+\mathrm{1}} }+… \\ $$$${t}−\frac{\mathrm{1}}{\mathrm{3}}{t}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{27}}+…. \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}{t}=\frac{\mathrm{1}/\mathrm{3}}{\mathrm{1}−\mathrm{1}/\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by necx122 last updated on 26/Oct/23
wow. This is a great solution. Thank you sir
Commented by Rasheed.Sindhi last updated on 26/Oct/23
Welcome necx sir, Ithink you are our old forum-friend!
$$\mathrm{Welcome}\:\mathrm{necx}\:\mathrm{sir},\:\mathrm{Ithink}\:\mathrm{you} \\ $$$$\mathrm{are}\:\mathrm{our}\:\mathrm{old}\:\mathrm{forum}-\mathrm{friend}! \\ $$
Commented by necx122 last updated on 26/Oct/23
Thank you. I'm shocked you still remember me. Truth is over these years (2014/2015 till now), I've learnt so much from you, Mr. W, Ajfour, Sir Abdo Imad, Alex342, 123456, Tanmay and my very good friend TawaTawa. Your solutions and help has helped me a long way. If there's one platform I'm most grateful about it's here, Tinkutara. There are no other great minds in Mathematics than here.��‍♂️

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *