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1-3-1-3-7-21-25-Next-three-terms-

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Question Number 215845 by Tawa11 last updated on 19/Jan/25
1, 3, − 1, − 3, − 7, − 21, − 25, ___, ___, ___ Next three terms??
$$\mathrm{1},\:\mathrm{3},\:−\:\mathrm{1},\:−\:\mathrm{3},\:−\:\mathrm{7},\:\:−\:\mathrm{21},\:\:−\:\mathrm{25},\:\:\:\:\:\_\_\_,\:\:\:\:\:\_\_\_,\:\:\:\:\:\_\_\_ \\ $$$$ \\ $$$$\mathrm{Next}\:\mathrm{three}\:\mathrm{terms}?? \\ $$
Commented by mr W last updated on 20/Jan/25
this is a game, not mathematics!!! the game is to guess what an other person thinks. mathematically you can put any three or thirty numbers and i can prove that they are right or that they are not right.
$${this}\:{is}\:{a}\:{game},\:{not}\:{mathematics}!!! \\ $$$${the}\:{game}\:{is}\:{to}\:{guess}\:{what}\:{an}\:{other} \\ $$$${person}\:{thinks}.\: \\ $$$${mathematically}\:{you}\:{can}\:{put}\:{any} \\ $$$${three}\:{or}\:{thirty}\:{numbers}\:{and}\:{i}\:{can} \\ $$$${prove}\:{that}\:{they}\:{are}\:{right}\:{or}\:{that}\:{they} \\ $$$${are}\:{not}\:{right}. \\ $$
Commented by mr W last updated on 20/Jan/25
i say three numbers: 1, 2, 3. now can you tell what are the next three numbers?
$${i}\:{say}\:{three}\:{numbers}:\:\mathrm{1},\:\mathrm{2},\:\mathrm{3}.\:{now}\: \\ $$$${can}\:{you}\:{tell}\:{what}\:{are}\:{the}\:{next}\:{three} \\ $$$${numbers}? \\ $$
Commented by Tawa11 last updated on 20/Jan/25
Sir, can this sequence be solved? If yes, please solve. a_(n + 1) = 3a_n ...... (i) a_(n + 1) = a_n − 4 .... (ii) Find a_n .
$$\mathrm{Sir},\:\mathrm{can}\:\mathrm{this}\:\mathrm{sequence}\:\mathrm{be}\:\mathrm{solved}? \\ $$$$\mathrm{If}\:\mathrm{yes},\:\mathrm{please}\:\mathrm{solve}. \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\mathrm{a}_{\mathrm{n}\:\:+\:\:\mathrm{1}} \:\:\:=\:\:\:\mathrm{3a}_{\mathrm{n}} \:\:\:\:\:\:\:\:……\:\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{a}_{\mathrm{n}\:\:+\:\:\mathrm{1}} \:\:\:=\:\:\:\mathrm{a}_{\mathrm{n}} \:\:−\:\:\mathrm{4}\:\:\:\:\:….\:\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{Find}\:\:\:\:\mathrm{a}_{\mathrm{n}} . \\ $$
Commented by mr W last updated on 20/Jan/25
you mean the question is a_0 =1 a_(n+1) =3a_n , if n is even a_(n+1) =a_n −4, if n is odd find a_n =?
$${you}\:{mean}\:{the}\:{question}\:{is} \\ $$$${a}_{\mathrm{0}} =\mathrm{1} \\ $$$${a}_{{n}+\mathrm{1}} =\mathrm{3}{a}_{{n}} ,\:{if}\:{n}\:{is}\:{even} \\ $$$${a}_{{n}+\mathrm{1}} ={a}_{{n}} −\mathrm{4},\:\:{if}\:{n}\:{is}\:{odd} \\ $$$${find}\:{a}_{{n}} =? \\ $$
Commented by Tawa11 last updated on 20/Jan/25
Yes sir. Exactly.
$$\mathrm{Yes}\:\mathrm{sir}.\:\mathrm{Exactly}. \\ $$
Commented by mr W last updated on 20/Jan/25
{ ((a_(2k) =2−3^k )),((a_(2k+1) =3(2−3^k ))) :}
$$\begin{cases}{{a}_{\mathrm{2}{k}} =\mathrm{2}−\mathrm{3}^{{k}} }\\{{a}_{\mathrm{2}{k}+\mathrm{1}} =\mathrm{3}\left(\mathrm{2}−\mathrm{3}^{{k}} \right)}\end{cases} \\ $$
Commented by mr W last updated on 20/Jan/25
we can also write in a single formula as a_n =3^(⌈(n/2)−⌊(n/2)⌋⌉) (2−3^(⌊(n/2)⌋) )
$${we}\:{can}\:{also}\:{write}\:{in}\:{a}\:{single}\: \\ $$$${formula}\:{as} \\ $$$${a}_{{n}} =\mathrm{3}^{\lceil\frac{{n}}{\mathrm{2}}−\lfloor\frac{{n}}{\mathrm{2}}\rfloor\rceil} \left(\mathrm{2}−\mathrm{3}^{\lfloor\frac{{n}}{\mathrm{2}}\rfloor} \right) \\ $$
Commented by Tawa11 last updated on 20/Jan/25
What does the bracket signify sir. And do you combine it sir.?
$$\mathrm{What}\:\mathrm{does}\:\mathrm{the}\:\mathrm{bracket}\:\mathrm{signify}\:\mathrm{sir}. \\ $$$$\mathrm{And}\:\mathrm{do}\:\mathrm{you}\:\mathrm{combine}\:\mathrm{it}\:\mathrm{sir}.? \\ $$
Commented by mr W last updated on 20/Jan/25
https://en.m.wikipedia.org/wiki/Floor_and_ceiling_functions
Commented by Tawa11 last updated on 20/Jan/25
Thanks sir.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Answered by Tawa11 last updated on 19/Jan/25
Commented by Tawa11 last updated on 19/Jan/25
1 × 3 = 3 first term by 3 to give second, then 3 - 4 = -1 to give the third. -1 × 3 = -3 (4th) -3 - 4 = -7 (5th) -7 × 3 = -21 etc
1 × 3 = 3 first term by 3 to give second, then
3 – 4 = -1 to give the third.
-1 × 3 = -3 (4th)
-3 – 4 = -7 (5th)
-7 × 3 = -21 etc
Answered by mr W last updated on 20/Jan/25
a_(2k+1) =3a_(2k) =3(a_(2k−1) −4)=3a_(2k−1) −12 let a_(2k+1) =Ap^(2k+1) +C Ap^(2k+1) +C=3Ap^(2k−1) +3C−12 A(p^2 −3)p^(2k−1) =2C−12 ⇒p^2 −3=0 ⇒p=±(√3) ⇒2C−12=0 ⇒C=6 a_(2k+1) =A((√3))^(2k+1) +6 a_1 =3a_0 =3×1=3=A((√3))^1 +6 ⇒A=−(√3) ⇒a_(2k+1) =−(√3)×((√3))^(2k+1) +6=6−3^(k+1) ⇒a_(2k+1) =3(2−3^k ) ⇒a_(2k) =2−3^k or a_(2k) =a_(2k−1) −4=3a_(2k−2) −4 let a_(2k) =Ap^(2k) +C Ap^(2k) +C=3Ap^(2k−2) +3C−4 A(p^2 −3)p^(2k−2) =2C−4 ⇒p^2 −3=0 ⇒p=±(√3) ⇒2C−4=0 ⇒C=2 a_(2k) =A((√3))^(2k) +2 a_0 =A((√3))^0 +2=1 ⇒A=−1 a_(2k) =2−3^k we can put both into a single formula a_n =3^(⌈(n/2)−⌊(n/2)⌋⌉) (2−3^(⌊(n/2)⌋) )
$${a}_{\mathrm{2}{k}+\mathrm{1}} =\mathrm{3}{a}_{\mathrm{2}{k}} =\mathrm{3}\left({a}_{\mathrm{2}{k}−\mathrm{1}} −\mathrm{4}\right)=\mathrm{3}{a}_{\mathrm{2}{k}−\mathrm{1}} −\mathrm{12} \\ $$$${let}\:{a}_{\mathrm{2}{k}+\mathrm{1}} ={Ap}^{\mathrm{2}{k}+\mathrm{1}} +{C} \\ $$$${Ap}^{\mathrm{2}{k}+\mathrm{1}} +{C}=\mathrm{3}{Ap}^{\mathrm{2}{k}−\mathrm{1}} +\mathrm{3}{C}−\mathrm{12} \\ $$$${A}\left({p}^{\mathrm{2}} −\mathrm{3}\right){p}^{\mathrm{2}{k}−\mathrm{1}} =\mathrm{2}{C}−\mathrm{12} \\ $$$$\Rightarrow{p}^{\mathrm{2}} −\mathrm{3}=\mathrm{0}\:\Rightarrow{p}=\pm\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}{C}−\mathrm{12}=\mathrm{0}\:\Rightarrow{C}=\mathrm{6} \\ $$$${a}_{\mathrm{2}{k}+\mathrm{1}} ={A}\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}{k}+\mathrm{1}} +\mathrm{6} \\ $$$${a}_{\mathrm{1}} =\mathrm{3}{a}_{\mathrm{0}} =\mathrm{3}×\mathrm{1}=\mathrm{3}={A}\left(\sqrt{\mathrm{3}}\right)^{\mathrm{1}} +\mathrm{6} \\ $$$$\Rightarrow{A}=−\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{a}_{\mathrm{2}{k}+\mathrm{1}} =−\sqrt{\mathrm{3}}×\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}{k}+\mathrm{1}} +\mathrm{6}=\mathrm{6}−\mathrm{3}^{{k}+\mathrm{1}} \\ $$$$\Rightarrow{a}_{\mathrm{2}{k}+\mathrm{1}} =\mathrm{3}\left(\mathrm{2}−\mathrm{3}^{{k}} \right) \\ $$$$\Rightarrow{a}_{\mathrm{2}{k}} =\mathrm{2}−\mathrm{3}^{{k}} \\ $$$${or} \\ $$$${a}_{\mathrm{2}{k}} ={a}_{\mathrm{2}{k}−\mathrm{1}} −\mathrm{4}=\mathrm{3}{a}_{\mathrm{2}{k}−\mathrm{2}} −\mathrm{4} \\ $$$${let}\:{a}_{\mathrm{2}{k}} ={Ap}^{\mathrm{2}{k}} +{C} \\ $$$${Ap}^{\mathrm{2}{k}} +{C}=\mathrm{3}{Ap}^{\mathrm{2}{k}−\mathrm{2}} +\mathrm{3}{C}−\mathrm{4} \\ $$$${A}\left({p}^{\mathrm{2}} −\mathrm{3}\right){p}^{\mathrm{2}{k}−\mathrm{2}} =\mathrm{2}{C}−\mathrm{4} \\ $$$$\Rightarrow{p}^{\mathrm{2}} −\mathrm{3}=\mathrm{0}\:\Rightarrow{p}=\pm\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}{C}−\mathrm{4}=\mathrm{0}\:\Rightarrow{C}=\mathrm{2} \\ $$$${a}_{\mathrm{2}{k}} ={A}\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}{k}} +\mathrm{2} \\ $$$${a}_{\mathrm{0}} ={A}\left(\sqrt{\mathrm{3}}\right)^{\mathrm{0}} +\mathrm{2}=\mathrm{1}\: \\ $$$$\Rightarrow{A}=−\mathrm{1} \\ $$$${a}_{\mathrm{2}{k}} =\mathrm{2}−\mathrm{3}^{{k}} \\ $$$${we}\:{can}\:{put}\:{both}\:{into}\:{a}\:{single}\:{formula} \\ $$$${a}_{{n}} =\mathrm{3}^{\lceil\frac{{n}}{\mathrm{2}}−\lfloor\frac{{n}}{\mathrm{2}}\rfloor\rceil} \left(\mathrm{2}−\mathrm{3}^{\lfloor\frac{{n}}{\mathrm{2}}\rfloor} \right) \\ $$
Commented by Tawa11 last updated on 20/Jan/25
Wow, thanks sir. It works for the sequence.
$$\mathrm{Wow},\:\mathrm{thanks}\:\mathrm{sir}. \\ $$$$\mathrm{It}\:\mathrm{works}\:\mathrm{for}\:\mathrm{the}\:\mathrm{sequence}. \\ $$
Commented by Tawa11 last updated on 23/Jan/25
Sir mrW, sorry for calling you everytime for my question. Please help on the mechanics question I posted sir. Thanks always sir. Q215975
$$\mathrm{Sir}\:\mathrm{mrW},\:\mathrm{sorry}\:\mathrm{for}\:\mathrm{calling}\:\mathrm{you}\:\mathrm{everytime}\:\mathrm{for} \\ $$$$\mathrm{my}\:\mathrm{question}.\:\mathrm{Please}\:\mathrm{help}\:\mathrm{on}\:\mathrm{the}\:\mathrm{mechanics}\:\mathrm{question} \\ $$$$\mathrm{I}\:\mathrm{posted}\:\mathrm{sir}.\:\mathrm{Thanks}\:\mathrm{always}\:\mathrm{sir}.\:\:\:\mathrm{Q215975} \\ $$

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