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Question Number 215845 by Tawa11 last updated on 19/Jan/25

$$1,3,-1,-3,-7,-21,-25,\mathrm{\_}\mathrm{\_}\mathrm{\_},\mathrm{\_}\mathrm{\_}\mathrm{\_},\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$$$$$\mathrm{Next}\mathrm{three}\mathrm{terms}??$$

Commented by mr W last updated on 20/Jan/25

$$thisisagame,notmathematics!!!$$$$thegameistoguesswhatanother$$$$personthinks.$$$$mathematicallyyoucanputany$$$$threeorthirtynumbersandican$$$$provethattheyarerightorthatthey$$$$arenotright.$$

Commented by mr W last updated on 20/Jan/25

$$isaythreenumbers:1,2,3.now$$$$canyoutellwhatarethenextthree$$$$numbers?$$

Commented by Tawa11 last updated on 20/Jan/25

$$\mathrm{Sir},\mathrm{can}\mathrm{this}\mathrm{sequence}\mathrm{be}\mathrm{solved}?$$$$\mathrm{If}\mathrm{yes},\mathrm{please}\mathrm{solve}.$$$$$$$${\mathrm{a}}_{\mathrm{n}+1}={3\mathrm{a}}_{\mathrm{n}}......\left(\mathrm{i}\right)$$$${\mathrm{a}}_{\mathrm{n}+1}={\mathrm{a}}_{\mathrm{n}}-4....\left(\mathrm{ii}\right)$$$$\mathrm{Find}{\mathrm{a}}_{\mathrm{n}}.$$

Commented by mr W last updated on 20/Jan/25

$$youmeanthequestionis$$$$a_0=1$$$$a_{n+1}=3a_n,ifniseven$$$$a_{n+1}=a_n-4,ifnisodd$$$$find{a}_n=?$$

Commented by Tawa11 last updated on 20/Jan/25

$$\mathrm{Yes}\mathrm{sir}.\mathrm{Exactly}.$$

Commented by mr W last updated on 20/Jan/25

$$\{\begin{array}{l}{a}_{2k}=2-3^k\\ a_{2k+1}=3(2-3^k)\end{array}$$

Commented by mr W last updated on 20/Jan/25

$$wecanalsowriteinasingle$$$$formulaas$$$$a_n=3^{\lceil \frac{n}{2}-\lfloor \frac{n}{2}\rfloor \rceil }(2-3^{\lfloor \frac{n}{2}\rfloor })$$

Commented by Tawa11 last updated on 20/Jan/25

$$\mathrm{What}\mathrm{does}\mathrm{the}\mathrm{bracket}\mathrm{signify}\mathrm{sir}.$$$$\mathrm{And}\mathrm{do}\mathrm{you}\mathrm{combine}\mathrm{it}\mathrm{sir}.?$$

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

$$\mathrm{Thanks}\mathrm{sir}.$$

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

Answered by mr W last updated on 20/Jan/25

$$a_{2k+1}=3a_{2k}=3(a_{2k-1}-4)=3a_{2k-1}-12$$$$let{a}_{2k+1}={Ap}^{2k+1}+C$$$${Ap}^{2k+1}+C=3{Ap}^{2k-1}+3C-12$$$$A(p^2-3)p^{2k-1}=2C-12$$$$\Rightarrow p^2-3=0\Rightarrow p=\pm \sqrt{3}$$$$\Rightarrow 2C-12=0\Rightarrow C=6$$$$a_{2k+1}=A{\left(\sqrt{3}\right)}^{2k+1}+6$$$$a_1=3a_0=3\times 1=3=A{\left(\sqrt{3}\right)}^1+6$$$$\Rightarrow A=-\sqrt{3}$$$$\Rightarrow a_{2k+1}=-\sqrt{3}\times {\left(\sqrt{3}\right)}^{2k+1}+6=6-3^{k+1}$$$$\Rightarrow a_{2k+1}=3(2-3^k)$$$$\Rightarrow a_{2k}=2-3^k$$$$or$$$$a_{2k}=a_{2k-1}-4=3a_{2k-2}-4$$$$let{a}_{2k}={Ap}^{2k}+C$$$${Ap}^{2k}+C=3{Ap}^{2k-2}+3C-4$$$$A(p^2-3)p^{2k-2}=2C-4$$$$\Rightarrow p^2-3=0\Rightarrow p=\pm \sqrt{3}$$$$\Rightarrow 2C-4=0\Rightarrow C=2$$$$a_{2k}=A{\left(\sqrt{3}\right)}^{2k}+2$$$$a_0=A{\left(\sqrt{3}\right)}^0+2=1$$$$\Rightarrow A=-1$$$$a_{2k}=2-3^k$$$$wecanputbothintoasingleformula$$$$a_n=3^{\lceil \frac{n}{2}-\lfloor \frac{n}{2}\rfloor \rceil }(2-3^{\lfloor \frac{n}{2}\rfloor })$$

Commented by Tawa11 last updated on 20/Jan/25

$$\mathrm{Wow},\mathrm{thanks}\mathrm{sir}.$$$$\mathrm{It}\mathrm{works}\mathrm{for}\mathrm{the}\mathrm{sequence}.$$

Commented by Tawa11 last updated on 23/Jan/25

$$\mathrm{Sir}\mathrm{mrW},\mathrm{sorry}\mathrm{for}\mathrm{calling}\mathrm{you}\mathrm{everytime}\mathrm{for}$$$$\mathrm{my}\mathrm{question}.\mathrm{Please}\mathrm{help}\mathrm{on}\mathrm{the}\mathrm{mechanics}\mathrm{question}$$$$\mathrm{I}\mathrm{posted}\mathrm{sir}.\mathrm{Thanks}\mathrm{always}\mathrm{sir}.\mathrm{Q}215975$$

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