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Re: HSC 2014 4U Marathon \sum_{k=1}^n k(k+1)(k+2)(k+3)=\\ \frac{1}{5}\sum_{k=1}^n \left[k(k+1)(k+2)(k+3)(k+4)-(k-1)(k)(k+1)(k+2)(k+3)\right]\\ = \frac{1}{5}n(n+1)(n+2)(n+3)(n+4).

Oh yeah I am also a sucker for canadian girls but I think that is more their typical personality than their accent. Still when I hear that accent on a girl it piques my interest.

On a girl: Certain kinds of english accents, spanish (and by extension some south american ones), russian to a lesser extent. On a guy: South African.

kit kats dipped in hommus. om nom nom.

Yeah, I assumed he meant for that to be 1/2 or -1/2. I didn't see the full progression of that working in my head but it seemed like it could work, although I think a bit slower than the method I posted. Would be interested in seeing if there was something quick he had in mind that I was missing.

I don't see it either, though you could possibly make something like that work.

Let S be this expression, use that cos(pi/9) = -cos(8pi/9), cos(6pi/9)=-1/2, and cos(pi+x)=cos(pi-x) to express the real part of: \sum_{k=0}^8 \textrm{cis}\left(\frac{2k\pi}{9}\right) in terms of S. What familiar polynomial is this the sum of roots of? z^9-1=0. This gives us a very simple...

Re: HSC 2014 4U Marathon Each term in the LHS sum is even, and the first term is 2. But 2 is the only even prime! So this sum cannot be prime.

First kiss was shit, second was amazing. A combination of the second girl being a lot better at kissing and both of us actually being into each other. Sidebar: The first kiss with someone new still hasn't lost its magic for me...I love the tension and the teasing.

Re: HSC 2014 4U Marathon Not quite, the logic is flawed at the start: we do not NEED a step of length n-1 in order to be maximal. Examples are not too tricky to find (the first example exists when n=6), but I think it might give a bit too much away if I provide one. The point is we don't need...

Re: HSC 2014 4U Marathon Yep, the main part of the problem is in determining the number of maximal trips (with proof) though.

Re: HSC 2014 4U Marathon Cute little combinatorics problem: A man is on the first floor of a building with n floors. He wants to use the elevator to stop at every floor exactly once. What are the minimum and maximum "lengths" of such a trip? How many trips of maximal length are there? You must...

In my experience, the vast majority of people who underrate it are people who haven't played through it. I too prefer Kefka but not by a very large margin. Most of the reason why I prefer ff6 to ff7 is that I prefer the characters, setting, and plot of ff6. It makes me actually care about...

Just started replaying my favourite game of all time (Final Fantasy 6 on the SNES), as I discovered a really nice rom modification that fixed bugs, changed mechanics, and upped difficulty (it is called Brave New World). If you have never played FF6 I strongly recommend it, this coming from a...

Pretty varied, depends on my mood. Mostly I just have pretty pure Ceylon. Ginseng green tea / other green tea is great on a cold evening. Stronger blends like Assam are good too. Am not that picky but I dislike earl grey and citrus/orange flavoured teas.

Wouldn't recommend it before an exam, having a good nights sleep and a solid breakfast is far more important. As part of your daily study routine, sure. I basically live off tea now, but coffee used to fill that niche for me.

Electric and acoustic guitar. Played the violin for ~7 years but haven't owned one for quite some time. Am looking to get back into the violin soon.

Re: HSC 2013 4U Marathon I can help out with any you are stuck with that aren't too ridiculous.

Haha the HS definitions are so hand-wavy and non-rigorous. You have to put them out of your mind when you are learning maths for real. The only kind of useful thing about high school maths is it will teach you to do certain computations quickly and develop a partial notion of rigor. To prove...

I meant why the word inquisitive? But yeah, the second method is the one I would use to rigorously develop arbitrary complex exponentiation. (Complicated by the fact that the logarithm is a multifunction on C).