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After trials you should have no gaps in your knowledge, if you still do quickly deal with it. Now all you need to do is improve problem solving skills/time management/accuracy Which is of course achieved by doing math papers, I recommend perhaps doing some of the older hsc papers, or if you...

Re: HSC 2013 4U Marathon Yep well done $Solve for all$ \ \ x \ $and$ \ y \frac{1}{x} + \frac{1}{y} = - 1 x^3+y^3 = 4

I'm not too concerned about English, its just that there are 2 heavy-content subjects right next to each other :/ And yes I did accelerate it

I'm not too sure, I think physics is on next Wednesday or something, not sure about other subjects. I'm just a little unlucky lol But I guess they need to leave room for marking and submission of marks to the board of studies.

Re: HSC 2013 4U Marathon $Prove$ \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots < 3

Good luck to you as well :)

Just when I thought it was over second set of trials starting tomorrow: timetable: Mon: Paper 1 Eng, Tues: Paper 2 Eng, Wed: EX1, Thursday: Chemistry, Friday: Economics Next Monday: MX2 Worst timetable

Re: HSC 2013 3U Marathon Thread Take the identity: (1+2x)^n = (x + (1+x))^n \sum_{k=0}^n \binom{n}{k} 2^k x^k = \sum_{k=0}^n \binom{n}{k} x^{n-k} (1+x)^k Then equate the co-efficients of x^k on both sides. =========================== $Prove by mathematical induction for...

I'll give you the first half of it, and let you finish it off. For b): It seems we need to somehow connect, a/b to b/a, b/c to c/b and c/a to a/c, so that we can get some 'cancellation' and get a definitive number. One may then proceed to add something to both sides but this doesn't look like...

Re: HSC 2013 4U Marathon Yep that is quite accurate of an estimate, the correct value is 302 Its better than my method though which was: (10^(log(2) * 1000) ), (using a calculator we get immediately that log(2) = 0.3016....) but my intention was for the answer to contain an attempt using...

you can send now, I've cleared some space.

Re: HSC 2013 4U Marathon $We want to find a sequence$ \ u_k $Such that$ \ \ \tan^{-1}(u_{k-1}) - \tan^{-1}(u_k) = \tan^{-1} \frac{1}{2k^2} From inspection, u_0 = 1 u_n = \frac{1}{2n+1} \therefore \ u_k = \frac{1}{2k+1} \ \ u_{k-1} = \frac{1}{2k-1} $We verify this by considering the...

Re: HSC 2013 4U Marathon Yep this was my method

Re: HSC 2013 4U Marathon I got: 1 - 1/n! + n/(n+1)! (only after a manipulation of my answer, I did not get it directly). What method did you use?

Re: HSC 2013 4U Marathon $Find$ \sum_{k=1}^n \frac{k}{(k+1)!}

I hate Labour markets definitely And things in topic 1 which is not protection Everything else I enjoy I think unless I'm forgetting something

Re: HSC 2013 4U Marathon A good Putnam problem that can be done with elementary knowledge $Evaluate$ \prod_{n=2}^{\infty} \frac{n^3-1}{n^3+1} \equiv \left( \frac{2^3-1}{2^3+1} \right ) \left ( \frac{3^3-1}{3^3+1}\right ) \left(\frac{4^3-1}{4^3+1} \right ) \left(\frac{5^3-1}{5^3 + 1}...

Re: HSC 2013 4U Marathon Well firstly I found the complementary probability, the probability that its closer to the face instead :/ And even then I still get the wrong answer because I don't know to find the area of a triangle properly and used the wrong base length. I think it should be...

Re: HSC 2013 4U Marathon Its just 4 times the area bounded by the region (in the Cartesian plane where the square's co-ordinates are (0,0) (1,0) (1,1) and (0,1) ) , y >x, \ \ y<(1-x) , (y-1/2)^2 > x-\frac{1}{4} ??

For a, consider the vector PA and rotate it 90 degrees to achieve the vector PO Let P represent the complex number p (p-z_1) cis(\frac{\pi}{2}) = (0-z_1) (p-z_1}i = (0-z_1) pi - z_1 i = -z_1 \therefore p = \frac{1}{i}z_1 (-1+i) = (1+i)z_1 b) We do something very...