HSC 2013 MX2 Marathon (archive) (4 Viewers)

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) (4 Viewers)

what is that?It is Cowpea

New Member

what is that?It is Cowpea

what is that?It is Cowpea

This too shall pass

Member

This too shall pass

New Member

This too shall pass

New Member

New Member

New Member

New Member

This too shall pass

Σ

This too shall pass

Σ

Member

Member

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 4)