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Question-186884

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Question Number 186884 by Mingma last updated on 11/Feb/23
Commented by mr W last updated on 11/Feb/23
you are asking questions non−stop, but you seem never to give any feed back: no thanks to people who answered your questions, no comments if the answers really have helped you etc... you have posted this question in Q 186657, and it is answered. what′s the reason that you posted the same question here again?
$${you}\:{are}\:{asking}\:{questions}\:{non}−{stop}, \\ $$$${but}\:{you}\:{seem}\:{never}\:{to}\:{give}\:{any}\:{feed} \\ $$$${back}:\:{no}\:{thanks}\:{to}\:{people}\:{who}\: \\ $$$${answered}\:{your}\:{questions},\:{no}\: \\ $$$${comments}\:{if}\:{the}\:{answers}\:{really} \\ $$$${have}\:{helped}\:{you}\:{etc}… \\ $$$$ \\ $$$${you}\:{have}\:{posted}\:{this}\:{question}\:{in}\: \\ $$$${Q}\:\mathrm{186657},\:{and}\:{it}\:{is}\:{answered}.\:{what}'{s} \\ $$$${the}\:{reason}\:{that}\:{you}\:{posted}\:{the}\:{same}\: \\ $$$${question}\:{here}\:{again}? \\ $$
Commented by Mingma last updated on 11/Feb/23
Bro, I give thumbs up when questions are answered. All questions you answered i gave thumbs up.
Commented by Mingma last updated on 11/Feb/23
Bro, I mistakenly posted it.
Commented by mr W last updated on 11/Feb/23
thanks for this feedback!
$${thanks}\:{for}\:{this}\:{feedback}! \\ $$
Commented by Mingma last updated on 11/Feb/23
Thank you too!
Commented by mr W last updated on 11/Feb/23
i prefer feedbacks in words. “likes”are no real comments, they are anonym and are often misused by some people.
$${i}\:{prefer}\:{feedbacks}\:{in}\:{words}.\: \\ $$$$“{likes}''{are}\:{no}\:{real}\:{comments}, \\ $$$${they}\:{are}\:{anonym}\:{and}\:{are}\:{often} \\ $$$${misused}\:{by}\:{some}\:{people}. \\ $$
Answered by ARUNG_Brandon_MBU last updated on 11/Feb/23
∗a_1 =7 , d=8 ⇒a_n =8n−1 ∗T_1 =3, T_(n+1) −T_n =a_n ⇒Σ_(k=1) ^n (T_(k+1) −T_k )=Σ_(k=1) ^n a_k ⇒T_(n+1) −T_1 =4n(n+1)−n=4n^2 +3n ⇒T_(n+1) =4n^2 +3n+3 ⇒T_n =4n^2 −5n+4 (i) Σ_(k=1) ^n T_k =((2n(n+1)(2n+1))/3)−((5n(n+1))/2)+4n =((4n^3 )/3)+2n^2 +((2n)/3)−((5n^2 )/2)−((5n)/2)+4n =((4n^3 )/3)−(n^2 /2)+((13n)/6) (ii) Σ_(k=1) ^n (T_(n+m) −T_k )=nT_(n+m) −(((4n^3 )/3)−(n^2 /2)+((13n)/6)) =4n(n+m)^2 −5n(n+m)+4n−(((4n^3 )/3)−(n^2 /2)+((13n)/6)) =((8n^3 )/3)+8n^2 m+4nm^2 −((9n^2 )/2)−5nm+((11n)/6) (iii)3T_(n+2) −3T_(n+1) +T_n =3(4(n+2)^2 −5(n+2)+4)−3(4(n+1)^2 −5(n+1)+4)+4n^2 −5n+4 =3(4n^2 +11n+10)−3(4n^2 +3n+3)+4n^2 −5n+4 =4n^2 +19n+25
$$\ast{a}_{\mathrm{1}} =\mathrm{7}\:,\:{d}=\mathrm{8}\:\Rightarrow{a}_{{n}} =\mathrm{8}{n}−\mathrm{1} \\ $$$$\ast{T}_{\mathrm{1}} =\mathrm{3},\:{T}_{{n}+\mathrm{1}} −{T}_{{n}} ={a}_{{n}} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({T}_{{k}+\mathrm{1}} −{T}_{{k}} \right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{k}} \\ $$$$\Rightarrow{T}_{{n}+\mathrm{1}} −{T}_{\mathrm{1}} =\mathrm{4}{n}\left({n}+\mathrm{1}\right)−{n}=\mathrm{4}{n}^{\mathrm{2}} +\mathrm{3}{n} \\ $$$$\Rightarrow{T}_{{n}+\mathrm{1}} =\mathrm{4}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}\:\Rightarrow{T}_{{n}} =\mathrm{4}{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{4} \\ $$$$\left({i}\right)\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{T}_{{k}} =\frac{\mathrm{2}{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{3}}−\frac{\mathrm{5}{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+\mathrm{4}{n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}{n}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2}{n}^{\mathrm{2}} +\frac{\mathrm{2}{n}}{\mathrm{3}}−\frac{\mathrm{5}{n}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{5}{n}}{\mathrm{2}}+\mathrm{4}{n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}{n}^{\mathrm{3}} }{\mathrm{3}}−\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{13}{n}}{\mathrm{6}} \\ $$$$\left({ii}\right)\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({T}_{{n}+{m}} −{T}_{{k}} \right)={nT}_{{n}+{m}} −\left(\frac{\mathrm{4}{n}^{\mathrm{3}} }{\mathrm{3}}−\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{13}{n}}{\mathrm{6}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}{n}\left({n}+{m}\right)^{\mathrm{2}} −\mathrm{5}{n}\left({n}+{m}\right)+\mathrm{4}{n}−\left(\frac{\mathrm{4}{n}^{\mathrm{3}} }{\mathrm{3}}−\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{13}{n}}{\mathrm{6}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{8}{n}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{8}{n}^{\mathrm{2}} {m}+\mathrm{4}{nm}^{\mathrm{2}} −\frac{\mathrm{9}{n}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{5}{nm}+\frac{\mathrm{11}{n}}{\mathrm{6}} \\ $$$$\left({iii}\right)\mathrm{3}{T}_{{n}+\mathrm{2}} −\mathrm{3}{T}_{{n}+\mathrm{1}} +{T}_{{n}} \\ $$$$\:\:\:\:\:\:\:=\mathrm{3}\left(\mathrm{4}\left({n}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{5}\left({n}+\mathrm{2}\right)+\mathrm{4}\right)−\mathrm{3}\left(\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{5}\left({n}+\mathrm{1}\right)+\mathrm{4}\right)+\mathrm{4}{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{4} \\ $$$$\:\:\:\:\:\:\:=\mathrm{3}\left(\mathrm{4}{n}^{\mathrm{2}} +\mathrm{11}{n}+\mathrm{10}\right)−\mathrm{3}\left(\mathrm{4}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}\right)+\mathrm{4}{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{4} \\ $$$$\:\:\:\:\:\:\:=\mathrm{4}{n}^{\mathrm{2}} +\mathrm{19}{n}+\mathrm{25} \\ $$
Commented by Mingma last updated on 11/Feb/23
Great work!
Commented by Mingma last updated on 12/Feb/23
I like the effort and demonstrations of the results. I wonder if a reduction of the third part could be done to bring the solution into a form recognizable in terms of T_{p}, where p = ?.

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