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Re: HSC 2014 4U Marathon Maybe this is better \\ $Find in the form$ \ (a+ib) \\ \\ $i)$ \ (1+i)(2+i) \\ \\ $ii)$ \ \frac{1+i}{2+i} \\ \\ $iii)$ (5+6i)+ (7+i) \\ \\ $iv)$ \ \frac{3+36i}{2+15i} \\ \\ $v)$ \ (6+i) - (\pi + \pi i) \\ \\ $vi)$ \ \frac{1+i}{1+i} \\ \\ $vii)$ \ \frac{1+12i}{4+3i} \\...

Re: HSC 2014 4U Marathon Indeed but the initial cis 7pi/12 provides a small obstacle to getting cis pi/12, so yes it makes it slightly necessarily harder

Re: HSC 2014 4U Marathon Not sure actually, depends on what you mean by hard, sketching weird implicit graphs can be seen as hard, such as: $Sketch$ \ \ x^2 - y^3 +xy -2y = 0

Re: HSC 2014 4U Marathon \\ $i) By process of Mathematical Induction show that$ \\ \\ (\cos x + i\sin x)^n = \cos(nx) + i\sin(nx) \ \ $for integer$ \ n \\ \\ $ii) Find$ \ x \ $in terms of$ \ \theta \ $such that$ \\ \\ 1 + x\cos \theta + x^2 \cos(2\theta) + x^3\cos(3\theta) + x^4\cos(4\theta) +...

Re: HSC 2014 4U Marathon \\ $i) Expand$ \ \ (1+i\sqrt{3})(1+i) \\ \\ $ii) Hence show that$ \ \ \cos \left(\frac{\pi}{24} \right) = \frac{1}{2} \sqrt{2 + \sqrt{2+\sqrt{3}}} If you want problems suited to a 2014er then I suggest posting some questions or answering the complex numbers that I...

Re: HSC 2014 4U Marathon a,b,c > 1 \frac{1}{a^2 -1} + \frac{1}{b^2 -1} + \frac{1}{c^2-1} = 1 \\ $Prove that$ \\ \\ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} \leq 1

Re: HSC 2014 4U Marathon How would a_1 + a_2 + \dots +a_n \geq na_1 Since a_1 is largest, the inequality sign should be reversed, meaning: \frac{1}{a_1} \leq n Which unfortunately cannot be allowed to 'be subbed in'

Re: HSC 2014 4U Marathon \\ $The sum of the positive reals$ \ a_1, a_2 , \dots , a_n \ $is$ \ 1 \\ \\ $Show that$ \\ \\ \left(1 + \frac{1}{a_1}\right) \left(1 + \frac{1}{a_2} \right) \left(1 + \frac{1}{a_3} \right) \dots \left(1+ \frac{1}{a_n} \right) \geq (n+1)^n

Re: HSC 2014 4U Marathon Correct! Though for a more rigorous solution I'd deal with the partial sum first but that's not much of an issue

Re: HSC 2014 4U Marathon Ahh yes I see that, editing.

Re: HSC 2014 4U Marathon \\ $Consider a set of$ \ n \ $distinct blue balls and$ \ n \ $distinct red balls. I will take any number of balls from the 2 boxes, that is, I can pick$ \ m \ $balls from the blue box and$ \ k \ $balls from the red box. The number of possibilities of such selections...

Re: HSC 2014 4U Marathon I like this sum: $Find the sum for$ \ |x| < 1 \frac{x}{1+x} + \frac{2x^2}{1+x^2} + \frac{4x^4}{1+x^4} + \dots

Re: HSC 2014 4U Marathon Well, lets just say its Harder 3U, where the 3U topic is Harder 2U

Re: HSC 2014 4U Marathon Strong quintuple post: $Sum the series$ 1\cdot 2 \cdot 3 \cdot 4 + 2\cdot 3 \cdot 4 \cdot 5 + 3 \cdot 4 \cdot 5 \cdot 6 + \dots + n \cdot (n+1) \cdot (n+2) \cdot (n+3)

Re: HSC 2014 4U Marathon I hope this is rigorous enough: $2) Inside the product, we will write each number in the form$ a= a_k 10^k + a_{k-1}10^{k-1} + \dots + a_1 10 + a_0 Hence our product is: 1\cdot 3 \cdot 7 \cdot 7 \cdot 9 \cdot (10 + 1) \cdot (10 + 3) \cdot (10+7) \cdot (10+9)...

Re: HSC 2014 4U Marathon $Draw a circle with chord$ \ AB \ $midpoint$ \ P \ $draw another chord$ \ CD \ $through$ \ P $Let$ \ \angle CAB = \angle CDB = \beta \angle ACD = \angle ABD = \alpha $Let$ \ AP = BP = a $By the sine rule$ \frac{CP}{\sin \beta} = \frac{a}{\sin \alpha}...

Re: HSC 2014 4U Marathon $1)$ \ 12^a \times 35^b \times 40^c = 2^{2a+3c} 5^{b+c} 3^a 7^b $Hence we must find the largest$ \ n \ $such that$ \frac{2^{2a+3c} 5^{b+c} 3^a 7^b}{2^{n} 5^n } = K \ $for integer$ \ K \therefore \ \ 2a+3c = n \ $and$ \ b+c \geq n $OR$ \ b+c = n \ $and$ \ 2a+3c...

Re: HSC 2014 4U Marathon Hmm I see..... In case of montone decreasing however, we simply need to see that: \sum_{k=1}^n \sqrt[k]{\frac{a_k}{a_{k+1}}} > \sum_{k=1}^n \sqrt[n]{\frac{a_k}{a_{k+1}}} Which would make a process identical to the montone increasing one Though for non-monotonic...

Re: HSC 2014 4U Marathon What do you mean?

Re: HSC 2014 4U Marathon Hopefully this works: $Without loss of generality, we assume the sequence$ \{a_k \} \ $from$ \ k=1, 2, \dots , n \ $is increasing$ $Therefore$ \ \ \frac{a_k}{a_{k+1}} < 1 $Therefore$ \ \ \frac{a_{k}}{a_{k+1}} < \sqrt[k]{\frac{a_{k}}{a_{k+1}}} \Rightarrow...