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Sorry - apologies. Fixed now.

What is the coefficient of x^5 and x^6? (3+2x)^n = \sum_{k=0}^n \binom{n}{k}3^{n-k}(2x)^k. Which value of k will give a power of x^5 (and hence x^6?)

Yep lel. Need to actually do some phys tmr.

Have you met Google?

Yep.

You should probably give a definition of inverse cotangent since it isn't explicitly defined in the syllabus.

omg LOL

The circle can be rotated exactly n ways which accounts for all the repeats. So if there are n seats, you need to do n! but divide by n, so n!/n = n(n-1)!/n = (n-1)!

Some of your working is incorrectly worded or deduced. You should say: LHS = 2^(3) - 3 = 8-3 = 5, which is divisible by 5, so the statement is true for n=1. By writing 5M immediately you are assuming that the left hand side is divisible by 5, you must deduce this - the fact that M=1 is...

Partial fractions.

Re: HSC 2013 4U Marathon Think of the subsets of \{ 1,2,\dots,n\} as picking 0,1,2,\dots elements from the set. That is, if you choose 0 elements, you have 1 choice. If you choose 1 element, you have n choices, for 2, nC2. Ie: \binom{n}{0} + \binom{n}{1}+\cdots+\binom{n}{n} = 2^n which is...

Re: HSC 2013 4U Marathon Think you meant: prove that x is a Fibonacci number if and only if 5x^2+4 or 5x^2-4 is a perfect square. (just being picky).

Let P be such a point with coordinates x and y. If it's distance from (1,-1) equals the distance from y=3...then... _________ = __________ (What's the distance between two general points?) (What's the distance between a point and y=3?)

Re: HSC 2013 4U Marathon Yep. So it'd just be -P(x)P(-x). I hope this is what sy was asking for.. Clearly that function is even, so we've checked that both plus or minus a_i is accounted for.

Re: HSC 2013 4U Marathon P(x) = k(x-a_1)(x-a_2)\cdots(x-a_n) \\ P(-x) = k(-x-a_1)(-x-a_2)\cdots(-x-a_n) Any takers?

O yep heh

Re: HSC 2013 4U Marathon Um... P(x) = (x-a_1)(x-a_2)\cdots(x-a_n)

Re: HSC 2013 4U Marathon It's actually simpler than that. A hint would be to use the root form of the polynomial. (Express it as a product of factors)

To be fair the question doesn't stipulate how many of the keys work so it's ambiguous

Draw a tree diagram - this will help.