Question

(Un) a sequence
U0=1 ; Un+1=(Un-1)/2
1) alpha is real number =/0
we have Vn=Un+alpha

What is the alpha value that make Vn geometric sequence

Answer 1

\[U_{0}=1 ; U_{n+1}=\frac{ U_{n}-1}{ 2}\]
\[V_{n}=U_{n}+alpha\]

Answer 2

For it to be a geometric sequence, it must be the case that
\[\large U_{n+1} = kU_n\]
For any nonzero constant, k.

Answer 3

what about our case

Answer 4

processing... this is tricky, lol

Answer 5

:) Please if you now something try to write it mathematically

Answer 6

Do you have to add alpha to U[0] as well?

Answer 7

no

Answer 8

the value of alpha to make vn geometric

Answer 9

So both sequences start at 1?

Answer 10

vn linked by Un , i think

Answer 11

@sirm3d what do you want to say

Answer 12

geometric sequence \(V_n\)

\[V_{n+1}=rV_n\]
substituting
\[U_{n+1}+\alpha=r(U_n+\alpha\\\frac{U_n-1}{2}+\alpha=rU_n+r\alpha\]

Answer 13

\[\frac{1}{2}U_n-\frac{1}{2}+\alpha=rU_n+r\alpha\]

therefore \[r=\frac{1}{2}\] and
\[-\frac{1}{2}+\alpha = \frac{1}{2}\alpha\] solving alpha, \[\alpha=1\]

Answer 14

this is @sirm3d you are the best

Answer 15

thank you

Answer 16

hang on...

Answer 17

Yeah, it works o.O

Answer 18

Thank you guys

Answer 19

yw

Answer 20

put \[\frac{V_{n+1}}{V_n}=\text{constant} = r\]

Answer 21

ok ok ok oko k

Answer 22

\[\frac{ U_{n}-1+2\alpha }{ 2(U_{n}+\alpha)}=C\]

Answer 23

If you let alpha by 1
you make un+1/2(un+1)=2 (constant)

Answer 24

clear the denominator first by writing it as \[U_n-1+2\alpha=2C(U_n+\alpha)\]
expand the right hand side, then compare the coefficients of \(U_n\)

Answer 25

C=1/2 ?

Answer 26

right.
use that value to solve \(\alpha\)

Answer 27

aaaaah yes

Answer 28

but it is dificult to solve like this equations ha ?

Answer 29

not really. the key to solving the equation is to compare coefficients of \(U_n\)

Answer 30

it is Conformity

Answer 31

and now how to write Vn terms n

Answer 32

The Vn term

Answer 33

write the terms of \(U_n\) and \(V_n\)
\[U_0=1, V_0=1+1=2\\U_1=\frac{1-1}{2}=0, V_1=0+1=1\\U_2=\frac{0-1}{2}=-\frac{1}{2},V_2=-\frac{1}{2}+1=\frac{1}{2}\\V_n=2\left(\frac{1}{2}\right)^n\]