HOLY SHIT WHAT IS THIS EPIPHANY OMFGn=2,, which is divisible by
Nuuuuuuuuuuuuuuuu wow I should've just wrote k=n+1
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- Thread starter Makematics
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Nuuuuuuuuuuuuuuuu wow I should've just wrote k=n+1
Makematics
Well-Known Member
bad luck!
anyways what i really wanna know is how you can actually prove this. Realise, i was also trying an induction in the last 5 minutes but to no avail.
RealiseNothing
what is that?It is Cowpea
My approach:bad luck!
Prove that:
Where 'p' is not divisible by 3.
Makematics
Well-Known Member
Everyone who did it seems to have found this question challenging.
}$ Suppose that $x_{1},x_{2},x_{3}$ and $y_{1},y_{2},y_{3}$ are positive real numbers such that$\\\\0< x_{1}y_{1}<x_{2}y_{2}<x_{3}y_{3}\\\\$and$\\\\x_{1}+x_{2}+x_{3}\geq y_{1}+y_{2}+y_{3}.\\\\$Prove that$\\\\\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\leq \frac{1}{y_{1}}+\frac{1}{y_{2}}+\frac{1}{y_{3}}.)
Makematics
Well-Known Member
Yep yep same here. I tried messing around with stuff like letting x=3^n but yeah, that didnt work out... was running out of time so just wrote down what i had observed from the results from the first few cases.My approach:
Prove that:
Where 'p' is not divisible by 3.
Makematics
Well-Known Member
Realise how did you do Q5?
Then factorise and equate, something along the lines of that. I had fricking n+1 the whole time and I kept stumbling with the "k=2 when n=2"
Everyone who did it seems to have found this question challenging.
OH YEAh that was a pretty sweet question
RealiseNothing
what is that?It is Cowpea
1) Prove that 'n' has to be a multiple of 2 (fairly straight forward and obvious as you need an even amount of terms so that all 1's and -1's cancel out).Realise how did you do Q5?
2)Prove that 'n' can be a multiple of 4. This is done by considering 4 terms:
Now this will always sum to 0 if only one of the x's is -1, and the other 3 are 1. So it is possible for 'n' to be a multiple of 4 by applying this method to all groups of 4.
3) Prove that 'n' can't be a multiple of 2 that isn't a multiple of 4, i.e.
Well you know the first
Makematics
Well-Known Member
i havent learnt harder 3u so my first reaction to Q4 was like "oh shiet, i need harder 3U skillz to do this"
Makematics
Well-Known Member
alright well i was messing around with cases and wrote something along those lines. someone who did it was saying something about factoring out terms and then summing them.1) Prove that 'n' has to be a multiple of 2 (fairly straight forward and obvious as you need an even amount of terms so that all 1's and -1's cancel out).
2)Prove that 'n' can be a multiple of 4. This is done by considering 4 terms:
Now this will always sum to 0 if only one of the x's is -1, and the other 3 are 1. So it is possible for 'n' to be a multiple of 4 by applying this method to all groups of 4.
3) Prove that 'n' can't be a multiple of 2 that isn't a multiple of 4, i.e.
Well you know the firstterms will sum to 0, so you prove the last 2 terms can never equal 0, etc.
RealiseNothing
what is that?It is Cowpea
I just realised for question 6, I could have written down the general formula:
which would have given me extra credit marks
which would have given me extra credit marks