Question

can you show me that Vn is a geometric progression with a common ratio 1/2?

Un+1=(3Un-1)/(2Un)
Vn=(Un-1)/(2Un-1)



i've already did this:
Vn+1=[(Un+1)-1]/[2(Un+1)-1]=[(3Un-1/2Un)-1]/[(2(3Un-1))/((2Un)-1)]=[3Un-1-2Un/2Un]/[2(3Un-1-2Un/2Un)]

but don't know what's next!! maybe

=[(3Un-1-2Un)2Un]/2[3Un-1-2Un)2Un)=[3Un-1-2Un]/[2(3Un-1-2Un)]

???????

Answer 1

you are doing way too much work

Answer 2

no matter what \(u_{n-1}\) is, it must be true that
\[\frac{u_{n-1}}{2u_{n-1}}=\frac{1}{2}\]

Answer 3

unless of course i am reading this wrong. hmm because it looks like \(v_n\) is identically \(\frac{1}{2}\) and not a geometric progression at all.
i am probably confused in identifying the subscripts and a subtraction maybe

Answer 4

yes you are confused :p anyway the problem is even with the answer i don't get it oO
in the second line how did we get [(4Un)-2] ??
Vn+1=[(Un+1)-1]/[2(Un+1)-1]
=[(Un)-1]/[(4Un)-2]
=(1/2)*[(Un)-1]/[(2Un)-1]
=1/2*Vn

Answer 5

\[V_{n+1} = [U_{n+1} -1]/[2U_{n+1}-1]\]\[= [[3U_{n} -1]/2U_{n} -1]/[2[3U_{n}-1]/2U_{n}]-1]\]\[=[[3U_{n}-1-2U_{n}]/2U_{n}]/[[3U_{n}-1-U_{n}]/U_{n}]\]\[=(1/2)[(U_{n}-1)/(2U_{n}-1)]\]\[=(1/2)V_{n}\]

Answer 6

@glgan1 Could you tell me this: [3Un−1−Un]/Un]?

Answer 7

\[2[(3U_{n}-1)/2U_{n}]-1\] \[=[(3U_{n}-1)/U_{n}]-1\]
then simplify it you will get the answer