really enjoyed this question, thought i'd share it so you can enjoy it as well.

i. Since all values \(x_n\) are positive, \(1/x_n\) is positive, so \[1+\frac{1}{x_n}\]is always greater than 1. ii. \[x_1-x_2=1+\frac{1}{x_2}-x_2=1+\frac{1}{x_2}-1-\frac{1}{x_3}\]\[=\frac{1}{x_2}-\frac{1}{x_3}=\frac{x_3-x_2}{x_2x_3}=-\frac{x_2-x_3}{x_2x_3}\]

i'm interested if you do part 3 the same way as me

iii. I'm trying to do a pure algebraic manipulation proof. I'm not sure if it's working yet. Since we have that that \(x_n=1+1/x_1\) we have that \[\Large x_n=1+\frac{1}{-\frac{x_2-x_3}{x_2x_3}+x_2}\]\[\Large =1+\frac{1}{-\frac{x_2-x_3+x_2x_2x_3}{x_2x_3}}\]\[\Large =1+\frac{{-x_2x_3}}{{x_2-x_3+x_2x_2x_3}}\]

my proof is different, i'll hold fire till you're done because yours looks interesting

Mine seems to have hit a brick wall however. Give me a few more minutes to see if I can get anywhere.

sure :D

Wait a second. I may have just missed the obvious. \[x_1=-\frac{x_2-x_3+x_2x_2x_3}{x_2x_3}\]so\[\frac{1}{x_1}=\frac{{-x_2x_3}}{{x_2-x_3+x_2x_2x_3}}\]We also know that \[x_n=\frac{1+x_1}{x_1}\]So using some substitutions...

\[|x_k-x_{k-1}|=\frac{1}{|x_{k+1}x_{k+2}|}\cdot |x_{k+1}-x_{k-2}|,k=1,2,\ldots,n-2\]\[|x_{n-1}-x_{n}|=\frac{1}{|x_{n}x_1|}\cdot |x_{n}-x_{1}|\]\[|x_{n}-x_{1}|=\frac{1}{|x_{1}x_2|}\cdot |x_{1}-x_{2}|\]Putting all this together we obtain:\[|x_1-x_2|=\frac{1}{M}\cdot |x_1-x_2|\]where M is:\[M=|x_1x_2\cdots x_n|^2\]Since each x_i is greater than 1, we have that:\[\frac{1}{M}<1\]Therefore:\[|x_1-x_2|=\frac{1}{M}|x_1-x_2|\Longrightarrow |x_1-x_2|=0\Longrightarrow x_1=x_2\]This can be done for any consecutive pair or numbers in the sequence.

\[\large x_n=\left(1+\frac{x_2-x_3+x_2x_2x_3}{-x_2x_3}\right)\left(\frac{{-x_2x_3}}{{x_2-x_3+x_2x_2x_3}}\right)\]If we distribute, we get \[\large x_n=\left(\frac{{-x_2x_3}}{{x_2-x_3+x_2x_2x_3}}+1\right)\]Nevermind. This just gets you back to where I was before. Joe's way actually seems to work.

my solution was along the same lines as joes i think: we use the result proven in part 2 \[x_1 - x_2 = \frac{x_2- x_3}{x_2 x_3}\] \[x_2 - x_3 = \frac{x_3- x_4}{x_3 x_4}\]...... \[x_n - x_1 = \frac{x_1- x_2}{x_1 x_2}\] the sum of these: \[0 = \sum_{r=1}^{n} \frac{x_r - x_{r+1}}{x_r x_{r+1}}\] so \[x_1 = x_2 = x_3 = x_4 = .... = x_n\]

part ii of the question leads to the conclusion that the differences between the consecutive terms are getting smaller and smaller, a contraction. But since this sequence cycles, that wouldnt make sense.

well now part ii just states 0 = 0

right. nothing makes sense, unless the terms are all equal.

I will still try to see if the algebraic form gets anywhere even though it's a terrible way to do it.

in my solution of part iii i missed my penultimate step of evaluating the series. ah well i think joe covered it in his :)

I can still do the last part however. Since all the terms are equal, \[x_1=1+\frac{1}{x_1}\]\[x_1^2-x_1-1=0\]Using the quadratic formula, we get that \[x_1=\frac{1+\sqrt{1-4(-1)}}{2}=\frac{1+\sqrt5}{2}=\varphi\]

We throw out the negative square root because \(x_1>1\).

yeaa

look at the file name ^

nice problem.

there's loads, if you want more: http://www.admissionstests.cambridgeassessment.org.uk/adt/step/Test+Preparation

they come in three difficulty levels, I , II , III there are mechanics and probability sections as well

while you're here is there an easier way than what i'm doing here: http://openstudy.com/study#/updates/4fc13702e4b0964abc832603

?

not 100% sure, but that problem feels like a Lagrange Multipliers Problem. You want to max/min the function:\[f(x,y,z)=(x-2)^2+(y-0)^2+(z-4)^2\]subject to the contraint:\[g(x,y,z)=\left(\frac{x}{5}\right)^2+\left(\frac{y}{2}\right)^2+\left(\frac{z}{10}\right)^2=1\]

that makes sense. do you know if the parametric thing i did would work?

ah nvm, i have to go. thanks for your help

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