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Re: HSC 2013 4U Marathon yeah, sorry for the lack of clarity.

Re: HSC 2013 4U Marathon yeah, sorry for the lack of clarity.

Re: HSC 2013 4U Marathon Exactly, the expression then becomes all cosines and the sine at the end is eliminated

Re: HSC 2013 4U Marathon Find the other expression for it using part i (just divide by theta then take n to infinity)

Re: HSC 2013 4U Marathon First define that 1+x > 0 therefore x > -1 Step 1: n=1 -> 1+x = 1+x Hence true for n=1 Step 2: Assumption Step 3: (1+x)^{k+1} \geq 1+(k+1)x \ \ \ ? From step 2: (1+x)^{k} (1+x) \geq (1+kx)(1+x) (1+x)^{k+1} \geq 1+ (k+1)x + kx^2 >...

Re: HSC 2013 4U Marathon $i) Show that$ \ \ \sin \theta = 2^{n} \cos (\theta /2) \cos (\theta /4) \cos (\theta /8) ... \cos (\theta / 2^n) \sin (\theta/ 2^n) \ \ \ \ \ \fbox{2} $ii) Hence show that$ \ \ \lim_{n \to \infty} \frac{\sin \theta}{\theta}=\cos...

I see, so it sort of balances out, the more tosses we make, the lower the probability, yet the more stuff we can add on so in the end it becomes 1 right? Though the sum of the probabilities is independent of n anyway...(obviously)

=========== Another thing: Ok, I want to find the probability of making n heads from 2n tosses of a coin. Using binomial probability: \binom{2n}{n} (1/2)^{2n} Now I can show via algebra, that this is largest value, hence it is most likely to have n heads n tails (this was meant to be the...

That's the problem. Its not fun.

Re: HSC 2013 3U Marathon Thread Nice work! I was aiming for someone to undergo the induction step: Step 3: n = k + 2 And if our initial condition is: Step 1: n= 2, then if we modify Step 3 to be such above then we can prove for all even integers. Similarly to prove something for all odd...

Re: HSC 2013 4U Marathon Ahaha, yeah my solution isn't as elegant as your one but I like it because it actually worked out. ========================= A question that is a little easier to other people don't get scared off or anything: x^3+3x+2=0 \ \ \ \ \rightarrow $roots$ \ \ \ \ \alpha...

Re: HSC 2013 4U Marathon Ok, I decided to undergo a proof by induction: $Step 1: $ \ \ n=4 If it is quadrilateral, it is obvious that there is only at most 1 point the diagonals can all intersect, and: 1=\binom{4}{4} Hence it is true for n=4 $Step 2:$ \ \ n=k We assume true that a k...

Re: HSC 2013 3U Marathon Thread $Prove that$ \ \ n(n+1)(n+2) \ \ $ is divisible by 12 for all EVEN integers n$ (yes it is very easy to prove without induction, but this question isn't meant for that)

Re: HSC 2013 4U Marathon Awesome question, loved it. So first of all, we must define the rotating circle equation: (x-h)^2+(y-k)^2 = 2.5^2 h, k are variable, they are variable with respect to the locus of the centre, and we are actually rotating the centre of the circle around the origin...

Re: HSC 2013 4U Marathon hahahahaha =============== I will post a new question soon though, I just need to verify the result for myself.

Re: HSC 2013 4U Marathon dont worry man, I scanned my solution up here you can check it if you want: http://www.ams.org/notices/199507/faltings.pdf

Re: HSC 2013 4U Marathon hey guys I made a new question: $i) Prove Goldbach's Conjecture$ \ \ \ \ \fbox{56} $ii) Prove the Hodge Conjecture$ \ \ \ \ \fbox{23} $iii) Prove Fermat's Last Theorem$ \ \ \ \ \fbox{442}

Re: HSC 2013 4U Marathon Freaking hell, why did I try and spend about an 1 hour trying to prove something that is out of the reach of even research mathematicians -.-

Re: HSC 2013 4U Marathon I initially thought of taking the case of n prime and n not prime and somehow using the prime number theorem thing. But I can't get anywhere. But we get 1000 000 for proving this then, I doubt the question is legit....

Re: HSC 2013 4U Marathon Nice work man. I will post another question soon (probably more about Palindromic polynomials), I first want to know what is fishy about the above question about the Harmonic numbers.