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

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

1, 3, - 1, - 3, - 7, - 21, - 25,___,___,___

Next three terms ? ?

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

t h i s i s a g a m e, n o t m a t h e m a t i c s!!!

t h e g a m e i s t o g u e s s w h a t a n o t h e r

p e r s o n t h i n k s .

m a t h e m a t i c a l l y y o u c a n p u t a n y

t h r e e o r t h i r t y n u m b e r s a n d i c a n

p r o v e t h a t t h e y a r e r i g h t o r t h a t t h e y

a r e n o t r i g h t .

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

i s a y t h r e e n u m b e r s : 1, 2, 3 . n o w

c a n y o u t e l l w h a t a r e t h e n e x t t h r e e

n u m b e r s ?

Commented by Tawa11 last updated on 20/Jan/25

Sir, can this sequence be solved ?

If yes, please solve .

a_{n + 1} = {3 a}_{n} \dots \dots (i)

a_{n + 1} = a_{n} - 4 \dots . (ii)

Find a_{n} .

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

y o u m e a n t h e q u e s t i o n i s

a_{0} = 1

a_{n + 1} = 3 a_{n}, i f n i s e v e n

a_{n + 1} = a_{n} - 4, i f n i s o d d

f i n d a_{n} = ?

Commented by Tawa11 last updated on 20/Jan/25

Yes sir . Exactly .

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

{\begin{cases} a_{2 k} = 2 - 3^{k} \\ a_{2 k + 1} = 3 (2 - 3^{k}) \end{cases}

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

w e c a n a l s o w r i t e i n a s i n g l e

f o r m u l a a s

a_{n} = 3^{⌈ \frac{n}{2} - ⌊ \frac{n}{2} ⌋ ⌉} (2 - 3^{⌊ \frac{n}{2} ⌋})

Commented by Tawa11 last updated on 20/Jan/25

What does the bracket signify sir .

And do you combine it 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 .

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_{2 k + 1} = 3 a_{2 k} = 3 (a_{2 k - 1} - 4) = 3 a_{2 k - 1} - 12

l e t a_{2 k + 1} = {A p}^{2 k + 1} + C

{A p}^{2 k + 1} + C = 3 {A p}^{2 k - 1} + 3 C - 12

A (p^{2} - 3) p^{2 k - 1} = 2 C - 12

\Rightarrow p^{2} - 3 = 0 \Rightarrow p = \pm \sqrt{3}

\Rightarrow 2 C - 12 = 0 \Rightarrow C = 6

a_{2 k + 1} = A {(\sqrt{3})}^{2 k + 1} + 6

a_{1} = 3 a_{0} = 3 \times 1 = 3 = A {(\sqrt{3})}^{1} + 6

\Rightarrow A = - \sqrt{3}

\Rightarrow a_{2 k + 1} = - \sqrt{3} \times {(\sqrt{3})}^{2 k + 1} + 6 = 6 - 3^{k + 1}

\Rightarrow a_{2 k + 1} = 3 (2 - 3^{k})

\Rightarrow a_{2 k} = 2 - 3^{k}

o r

a_{2 k} = a_{2 k - 1} - 4 = 3 a_{2 k - 2} - 4

l e t a_{2 k} = {A p}^{2 k} + C

{A p}^{2 k} + C = 3 {A p}^{2 k - 2} + 3 C - 4

A (p^{2} - 3) p^{2 k - 2} = 2 C - 4

\Rightarrow p^{2} - 3 = 0 \Rightarrow p = \pm \sqrt{3}

\Rightarrow 2 C - 4 = 0 \Rightarrow C = 2

a_{2 k} = A {(\sqrt{3})}^{2 k} + 2

a_{0} = A {(\sqrt{3})}^{0} + 2 = 1

\Rightarrow A = - 1

a_{2 k} = 2 - 3^{k}

w e c a n p u t b o t h i n t o a s i n g l e f o r m u l a

a_{n} = 3^{⌈ \frac{n}{2} - ⌊ \frac{n}{2} ⌋ ⌉} (2 - 3^{⌊ \frac{n}{2} ⌋})

Commented by Tawa11 last updated on 20/Jan/25

Wow, thanks sir .

It works for the 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 . Q 215975

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