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group 2 E's, do prob as though they are 1 letter, then subtract the answer from 14a basically 2 E's together is inclusive of 3 E's together so you can subtract it.

Re: HSC 2014 4U Marathon part 4 is essentially the same as part 1. separate a1, a3, a5 etc. then assume that a1+ a3+ a5 etc = a2 + a4 + a6 + etc. you can factorise out an 11 from there. for the difference bit, you just add a2+a4+a6+... to the series and factor out an 11, in the process you...

basically if those are the prime factors, you can include any of them in a divisor raised to any integer power, maximum being the index of the prime factors, but also including ^0. so an example of a diviser would be 2^0 * 3^2 * 5^3 * 7^0 then it's just fundamental counting theory multiply the...

Re: HSC 2014 4U Marathon - Advanced Level got it from a mate. and i honestly have no idea

Re: HSC 2014 4U Marathon - Advanced Level I was hoping you guys could find one hence why i posted it here. i spent a couple of days on it couldn't get anything :P

Re: HSC 2013 3U Marathon Thread For part ii) \\ S(1) = 0, S'(1) = 0, S(-1) = 4 \\ a+b+c = 0 \quad (1) \\ c-a-b = 4 \quad (2) \\ 3a + b = 0 \quad (3) \\ (1)+(2) \\ c = 2 \\ a+b = -2 \\ 2a - 2 = 0 \\ a = 1 \\ b = -3 \\ S(y) = y^3 -3y + 2

Re: HSC 2013 3U Marathon Thread mathematically, you can just differentiate it.

Re: HSC 2014 4U Marathon Yes sorry, typo :P My hand written version included the diagram :P

Re: HSC 2014 4U Marathon oh true yeah. derp moment T_T

Re: HSC 2014 2U Marathon well i think for part 2 you were supposed to use part 1. its pretty simple. also induction is a 3U technique, this is a 2U thread :)

Re: HSC 2014 4U Marathon but it's still not proven for all odd, just shown for the first 4

Re: HSC 2013 3U Marathon Thread \text{Testing for n=3, we get a triangle, and we know that the angle sum of a triangle is } \pi \\ \text{Therefore, statement is true for n=3} \\ \text{Assuming to be true for n=k} \\ \text{i.e. Angle sum of k sided polygon is given by } \pi(k-2) \\...

Re: HSC 2013 3U Marathon Thread are we allowed to make the assumption that the angle sum of a triangle is pi?

Re: HSC 2014 4U Marathon equivalent of letting x = 2n-1, where n is any integer. let x = 2n-1 where n is an integer. \text{It is required to prove that }(2n-1)^{2} - 1 = 8M \text{ for all natural numbers n, where M is an integer} \\ \text{Test n=1, statement is true} \\ \text{Assume true for...

Re: HSC 2014 4U Marathon 4=4 \\ -21 = -21 \\ 9-30 = 49-70 \\ 9-30+25 = 49-70+25 \\ (3-5)^2 = (7-5)^2 \\ 3-5 = 7-5 \\ 1+2 = 7 gg

Re: HSC 2014 4U Marathon Gonna skip lots of lines, especially in the earlier parts, because this is extremely long. \\ (i) \quad u=(1-lnx) \quad dv = dx \quad I_{n} = [x(1-lnx)^{n}]_{1}^{e} + n\int_{1}^{e}(1-lnx)^{n-1}dx \\ I_{n} = -1 + nI_{n-1} \\ \\ (ii) \quad I_{3} = 6e - 16 \\ \\ (iii)...

Re: HSC 2014 4U Marathon - Advanced Level Well assuming they are both perfect squares of integers a & b, their product would be a perfect square of ab. So you would be left attempting to show that ab is not at integer.

Re: HSC 2013 3U Marathon Thread \\ \text{Since PQRS is cyclic, } \angle{PQR} = 180-\theta \\ \text{By use of the cosine rule in triangle PQR, } \\ (PR)^{2} = a^{2} + c^{2} - 2accos(180-\theta) = a^{2} + c^{2} + 2accos\theta \\ \text{By use of the sine rule in triangle PRS, }...

Re: HSC 2013 3U Marathon Thread spent 10 minutes trying to figure this out then noticed the question said CYCLIC. fml

Re: HSC 2014 4U Marathon - Advanced Level It can be expressed in that form though. (\sqrt{17m^{2} + 4n^{2}} +\sqrt{17n^{2} + 4m^{2}} )(\sqrt{17m^{2} + 4n^{2}} - \sqrt{17n^{2} + 4m^{2}} ) You would have to then show that a and b cant both be integers.