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Re: HSC 2015 4U Marathon - Advanced Level Damn, difficult question. I agree with you up to about the final line. By substituting x = -k, we obtain the constants: A_k = \frac{1}{(-1)^{k}(n-k)!k!} So the final answer should be: \ln\prod_{k=0}^n (x+k)^{\frac{1}{(-1)^{k}(n-k)!k!}} \, + C...

Re: HSC 2015 4U Marathon - Advanced Level Apply the AP-GP inequality for integers 1 to n: \frac{1}{n}\cdot\frac{n(n+1)}{2}\geq\sqrt[n]{n!} Making the cancellation and moving the root to the other side, we obtain the result.

4x^4 - 4x^2 - 9x^2 +9 = 4x^2(x^2 - 1) - 9(x^2 -1) \\= (4x^2 - 9)(x^2 - 1) = (2x+3)(2x-3)(x+1)(x-1)

4320 is wrong, it's short by a factor of 7 (and I think it assumes there are 6 distinct groups being arranged, rather than 7). Your answer's right.

and the agreed-on meeting place? :)

I vote Tuesday, Wednesday and Thursday (did the polls get my vote?)

Re: HSC 2015 4U Marathon - Advanced Level I'm a bit rusty, but let's give this a go. Consider each student's report in terms of an 8-element, ordered string of A's and B's. We know we must satisfy both the stated conditions, that (1) no two strings are the same, and (2) there cannot be a...

gg mreditor 99.60

Re: Not gettin SR with 100 overall In my opinion, the SR list will tell. A lot of people are having the same problem with marks in the 98-100 range without any indication of an SR.

Re: HSC 2015 4U Marathon - Advanced Level Sy, empty your PM box

Re: HSC 2015 4U Marathon - Advanced Level Alternatively, set: p(x) = \left[x-(p+\sqrt{q})\right]\left[x+(p-\sqrt{q})\right]\cdot q(x)+ax+b =\left[(x-p)^{2} - q\right]\cdot q(x) +ax+b where the coefficients of q(x), together with a and b, are rational. Since p+\sqrt{q} is a root...

Re: HSC 2015 4U Marathon - Advanced Level For non-zero p: If p(x) has integer coefficients, then the product of its roots must be of a rational form (the quotient of two integers, the leading coefficient and the constant). To rationalize such a product with some irrational p+\sqrt{q} we...

Re: HSC 2015 4U Marathon - Advanced Level Denote the summation on the LHS as S so that: $$2S = \frac{3}{4} +\frac{4}{8} + \frac{5}{16} + \dots$$ By subtraction: S = \frac{3}{4} + \left(\frac{4}{8}-\frac{3}{8}\right) +...

Re: HSC 2014 4U Marathon - Advanced Level Sum the components of the product and divide it by the number of terms so that the geometric mean is less than the arithmetic one: \\ \frac{(1+m)+1+1+1+...}{n} Notice we are adding n-1 1's to 1+m, (the latter counting for 1 term).

Re: HSC 2014 4U Marathon - Advanced Level woah what the hell am I doing, wishful thinking. sorry guys, ignore that

Re: HSC 2014 4U Marathon - Advanced Level Drop perpendiculars from U and Z to XY. From a consideration of the right angled triangles formed thereby, it should be easy trigonometry that XU = (sin\,6^{\circ})^{-1}, YZ = (sin\,42^{\circ})^{-1}, UB = (sin\,78^{\circ})^{-1} and ZB =...

Re: HSC 2014 4U Marathon - Advanced Level My bad. Question's asked incorrectly then. ftfy

Re: HSC 2014 4U Marathon - Advanced Level Also I think the third line should read H_n - \ln n.

Re: HSC 2014 4U Marathon - Advanced Level Not really. Investigating monotonic behaviour in functions and looking at boundaries is well within the scope of Extension 2. Consider a_n = H_n - \ln n. So: a_n - a_{n-1} = \frac{1}{n} - \ln n + \ln (n-1) \\= \frac{1}{n}+\ln...

The "conversion" to which you refer has been cited incorrectly. In the snapshot (Cambridge, from the font I think?) there is a cube on both sides. What's inside the brackets on each side should be clear through a re-arrangement of 1+ {\omega} + {\omega}^2 = 0 . Move the squared term to the...