OpenStudy (anonymous):

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)] ???????

5 years ago
OpenStudy (anonymous):

you are doing way too much work

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
Similar Questions: