HSC 2013 MX2 Marathon (archive) (1 Viewer)

RealiseNothing · Feb 3, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
yeah he has (should have done the HSC last year) lol

This, I accelerated 4U in year 11 but teacher wouldn't let me do the HSC exam

(despite getting 95/100 raw in trials he didn't think I was ready... yer idk)

Helvetica · Feb 3, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
This, I accelerated 4U in year 11 but teacher wouldn't let me do the HSC exam (despite getting 95/100 raw in trials he didn't think I was ready... yer idk)

WOW!!

RealiseNothing · Feb 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Prove that$ \ \ \frac{a+b}{2} \geq \sqrt{ab}$

$Using the above result, hence prove that$

$\frac{a+b+c+d}{4} \geq \sqrt[4]{abcd}$

You can use the AM-GM inequality (the first result) to prove a really nice circle geo theorem/result. I've made it a question but I'm saving it for later though (going to write my own trials if I have time).

RealiseNothing · Feb 3, 2013

Re: HSC 2013 4U Marathon

Helvetica said:
WOW!!

Admittedly it was an incredibly easy test lol.

Sy123 · Feb 3, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
You can use the AM-GM inequality (the first result) to prove a really nice circle geo theorem/result. I've made it a question but I'm saving it for later though (going to write my own trials if I have time).

I want to do this but I don't want to let the marathon die lol (even though its on the verge of death at the moment)

================

$Find rational$ \ \ x \ \ $such that$

$\sqrt{x} = \sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}}$

Lieutenant_21 · Feb 3, 2013

Re: HSC 2013 4U Marathon

x=2+sqrt3+2(1)+2-sqrt3=6

Sy123 · Feb 3, 2013

Re: HSC 2013 4U Marathon

$i) Show that$

$1+2z+3z^2+\dots+nz^{n-1} = \frac{nz^{n+1}-(n+1)z^n +1}{(1-z)^2} \ \ \ |z|<1 \ \ \ \fbox{2}$

$$ii) Hence find in terms of $ \ z \ \ \ \ \sum_{k=1}^{\infty} kz^k \ \ \ \fbox{1}$

$iii) The Triangular numbers are a sequence of numbers defined by the recursive relations$

$T_1=1 \ \ \ T_k=k+T_{k-1}$

$Define the power series$ \ \ T(z) \ \ $, where the co-efficents are the triangular numbers$

$T(z) = \sum_{k=1}^{\infty} T_k z^k \equiv z + 3z^2+6z^4+10z^3+...$

$By using the recursive definition of$ \ \ T_k \ \ $ and using part ii, prove that$

$T(z) = \frac{z}{(1-z)^3} \ \ \ \ \fbox{4}$

Omed69 · Feb 3, 2013

Re: HSC 2013 4U Marathon

Helvetica said:
$Induction Question$
$Prove:$ $(2^{b}-1).\sum_{k=0}^{a-1}2^{kb}=2^{ab}-1$

This is a very simple 3U induction question, why did you post it here? I will show you how to do it. I promise you guys I will keep the thread alive. Realise and Si helped me a lot so this is me returning the favour.
$Base case: When a=1 we have:$ $(2^{b}-1).\sum_{k=0}^{a-1}2^{kb}=(2^{b}-1).\sum_{k=0}^{0}2^{kb} = 2^{b}-1$
$$Now suppose$ $a\geq 1 and (2^{b}-1).\left ( \sum_{k=0}^{a-1}2^{kb}+2^{ab} \right ) = (2^{b}-1).\sum_{k=0}^{a-1}2^{kb}+2^{b}.2^{ab}-2^{ab} = 2^{ab}-1+2^{b+ab}-2^{ab} = 2^{(a+1)b}-1. \square$

Sy123 · Feb 3, 2013

Re: HSC 2013 4U Marathon

Omed69 said:
Question:

Are you sure that isn't the answer?

Omed69 · Feb 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Are you sure that isn't the answer?

I thought the question is too difficult for students undertaking 4 units on maths so I posted the solution.

Omed69 · Feb 3, 2013

Re: HSC 2013 4U Marathon

Here is a question for you Si123:

Omed69 · Feb 3, 2013

Re: HSC 2013 4U Marathon

He is another very beaiutiful question for you guys:

Omed69 · Feb 3, 2013

Re: HSC 2013 4U Marathon

Hi,
Its me Omed,
You have not solved my questions yet. You are disappointing.

Sy123 · Feb 3, 2013

Re: HSC 2013 4U Marathon

Omed69 said:
Hi,
Its me Omed,
You have not solved my questions yet. You are disappointing.

Relax I only got to have time to think of your problem now, its actually very simple:

Omed69 said:
Here is a question for you Si123:

$a \leq 1$
$a+b \leq 5$
$a+b+c \leq 14$
$a+b+c+d \leq 30$

Now, using subtraction, it follows that:

$a\leq 1 \ \ b \leq 4 \ \ c \leq 9 \ \ d \leq 16$

Conveniently

$\sqrt{a} \leq 1 \ \ \sqrt{b} \leq 2 \ \ \sqrt{c} \leq 3 \ \ \sqrt{d} \leq 4$

Add all together:

$\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} \leq 10$

And your second question is out of curriculum

asianese · Feb 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$i) Show that$

$1+2z+3z^2+\dots+nz^{n-1} = \frac{nz^{n+1}-(n+1)z^n +1}{(1-z)^2} \ \ \ |z|<1 \ \ \ \fbox{2}$

$$ii) Hence find in terms of $ \ z \ \ \ \ \sum_{k=1}^{\infty} kz^k \ \ \ \fbox{1}$

$iii) The Triangular numbers are a sequence of numbers defined by the recursive relations$

$T_1=1 \ \ \ T_k=k+T_{k-1}$

$Define the power series$ \ \ T(z) \ \ $, where the co-efficents are the triangular numbers$

$T(z) = \sum_{k=1}^{\infty} T_k z^k \equiv z + 3z^2+6z^4+10z^3+...$

$By using the recursive definition of$ \ \ T_k \ \ $ and using part ii, prove that$

$T(z) = \frac{z}{(1-z)^3} \ \ \ \ \fbox{4}$

The top sum is incorrect I think...

Sy123 · Feb 3, 2013

Re: HSC 2013 4U Marathon

asianese said:
The top sum is incorrect I think...

Are you sure? Wolfram Alpha says it is correct

asianese · Feb 3, 2013

Re: HSC 2013 4U Marathon

No wait nvm I typed into wolfram wrong :\

Lieutenant_21 · Feb 3, 2013

Re: HSC 2013 4U Marathon

Omed69 said:
He is another very beaiutiful question for you guys:

I have seen this question in the French maths Olympias IIRC. Did you get it from there?

Lieutenant_21 · Feb 3, 2013

Re: HSC 2013 4U Marathon

Off-topic: http://www.youtube.com/watch?v=WGe-nGZLENU
Jealous

Sy123 · Feb 4, 2013

Re: HSC 2013 4U Marathon

$A sequence is defined recursively as follows$

$F_0 = 0 \ \ F_1=1$

$F_n=F_{n-1}+F_{n-2} \ \ \ n \geq 2$

$$i) Write down the first 7 numbers of this sequence$ \ \ \ \fbox{1}$

$ii) Consider a power series in$ \ z \ $whereby the co-efficients are the numbers of the defined sequence. That is$

$F(z)=\sum_{k=0}^{\infty} F_k z^k \equiv z+z^2+2z^3+3z^4+5z^5+8z^6+ \dots$

$$Show that$ \ \ \ F(z)=z+zF(z)+z^2F(z) \ \ \ \fbox{2}$

$$iii) Hence show that$ \ \ \ F(z) = \frac{-z}{z^2+z-1} \ \ \ \ \fbox{1}$

$iv) Find$ \ \ A, \ B, \ \alpha, \ \beta \ \ $within the below expression$

$F(z)=\frac{A}{1-\frac{z}{\alpha}} + \frac{B}{1-\frac{z}{\beta}}$

$$Where$ \ \ A \ \ $and$ \ \ B \ \ $are independent of$ \ \ z \\ \\ $ and$ \ \ \alpha, \ \beta \ \ $are roots of$ \ \ \ z^2+z-1 \ \ \ \ \fbox{2}$

$v) Hence prove that$

$F_k = \frac{1}{\sqrt{5}} \left ( \left(\frac{1+\sqrt{5}}{2} \right )^k - \left (\frac{1-\sqrt{5}}{2} \right )^k \right ) \ \ \ \fbox{2}$

HSC 2013 MX2 Marathon (archive) (1 Viewer)

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