HSC 2013 MX2 Marathon (archive) (3 Viewers)

hamstar · May 17, 2013

Re: HSC 2013 4U Marathon

I think the answer is

(n+1)! - 1

you can even prove by mathematical induction.

nightweaver066 · May 17, 2013

Re: HSC 2013 4U Marathon

asianese said:
Methinks he meant

$\lim_{ n \to\infty} \sum_{k=0}^n k\times k!$

naw i don't think so.

(n+1)! - 1

edit: lol woops, haven't refreshed for a while

Registered User · May 17, 2013

Re: HSC 2013 4U Marathon

hamstar said:
I think the answer is

(n+1)! - 1

you can even prove by mathematical induction.

This is right.
The whole point of the question is to get a telescoping sum... those come up in the HSC a lot (not really).

Sy123 · May 18, 2013

Re: HSC 2013 4U Marathon

$Prove without using the Binomial Theorem$

$\binom{2n}{n} < 4^n$

seanieg89 · May 18, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove without using the Binomial Theorem$

$\binom{2n}{n} > 4^n$

Err...

seanieg89 · May 18, 2013

Re: HSC 2013 4U Marathon

bleakarcher said:
Hey guys, here's a question from last year's UNSW maths competition. Let f(x) denote a strictly positive continuous function defined on all real numbers with the property that f(2012)=2012 and f(x)=f(x+f(x)) for all x. Prove that f(x)=2012 for all x.

Tricky question! Am clearly rusty at contest mathematics.

$By a straightforward induction we get $f(g_n(x))=f(x)$ for all $x$, where $g_n(x)=x+nf(x)$. Now we claim that each $g_n$ is injective.\\ $g_n(x)=g_n(y)\Rightarrow f(x)=f(g_n(x))=f(g_n(y))=f(y)\Rightarrow x=g_n(x)-nf(x)=g_n(y)-nf(y)=y.$\\ \\ As continuous injections are monotone (a fact easy to prove via the intermediate value theorem) and each $g_n$ tends to infinity as $x\rightarrow \infty$, we get that each $g_n$ is strictly increasing on $\mathbb{R}$.\\ \\Hence if we fix $x<y$ we get $x+nf(x)<y+nf(y)$ for all positive integers $n$. This will fail to be true for sufficiently large $n$ if $f(x)>f(y)$, and hence our sought function $f$ must be non-decreasing.\\ \\As $f$ is non-decreasing and for every $x$ and every $m$ we have $f(x)=f(x+mf(x))$, we get that $f$ is constant on the interval $[x,\infty)$ for every $x$. \\ \\Hence $f$ is constant on the entire real line and $f(x)=2012$ for all $x$.$

Sy123 · May 18, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Err...

Oops heh, fixed.

Sy123 · May 18, 2013

Re: HSC 2013 4U Marathon

$$i) Prove that$ \ \ \int_{0}^{\frac{\pi}{2}} \sin^{2n-1} x \ dx = \frac{2 \cdot 4 \cdot 6 \dots (2n-2)}{1 \cdot 3 \cdot 5 \dots (2n-1)} \ \ \ \fbox{2}$

$ii) Hence or otherwise prove that$

$\sum_{n=1}^{\infty} \frac{1}{\binom{2n}{n}} = \frac{1}{3} + \frac{2\pi \sqrt{3}}{27} \ \ \ \ \fbox{6}$

RealiseNothing · May 18, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove without using the Binomial Theorem$

$\binom{2n}{n} < 4^n$

I'm not sure if this counts as using the Binomial Theorem or not, but consider the following:

$\sum_{k=0}^{2n} \binom{2n}{k} = 2^{2n} = 4^n$

Then the result falls out immediately.

Sy123 · May 18, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
I'm not sure if this counts as using the Binomial Theorem or not, but consider the following:

$\sum_{k=0}^{2n} \binom{2n}{k} = 2^{2n} = 4^n$

Then the result falls out immediately.

You could do that, but of course there are ways of proving that result without using Binomial theorem (i.e. combinatorics), iirc one of your questions on here was about that.

RealiseNothing · May 18, 2013

Re: HSC 2013 4U Marathon

realise you are a fucking idiot

asianese · May 18, 2013

Re: HSC 2013 4U Marathon

d....x?

Trebla · May 19, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove without using the Binomial Theorem$

$\binom{2n}{n} < 4^n$

Assuming n is a positive integer this can be easily proved by induction.

Alternatively:

$\binom{2n}{n}\\\\=\dfrac{(2n)!}{n!\times n!}\\\\=\dfrac{1\times 3\times 5\times ..... \times (2n-1)}{1\times 2\times 3\times ..... \times n} \times \dfrac{2 \times 4 \times 6 \times ..... \times (2n)}{1\times 2 \times 3 \times ..... \times n}\\\\ =2^n\displaystyle\prod_{k=1}^n\dfrac{2k-1}{k}\\\\ < 2^n \displaystyle\prod_{k=1}^n\dfrac{2k}{k} \\\\ = 4^n$

Trebla · May 19, 2013

Re: HSC 2013 4U Marathon

I got a feeling I've asked this before, but will put it in anyway.

Prove that for any set of real numbers {x_i}

$\displaytsyle\sum_{i=1}^n x_i^2 \geq \dfrac{1}{n}\left(\displaystyle\sum_{i=1}^n x_i \right)^2$

Also, here is a Conics question for the purposes of variety on this thread:

The points P(a cos θ, b sin θ) and Q(a cos φ, b sin φ) lie on an ellipse. If PQ is a focal chord which passes through the positive focus show that the length of PQ is

$2a\left(1 - e \cos\dfrac{\theta + \psi}{2}\right)\left(1 + e \cos\dfrac{\theta + \psi}{2}\right)$

where e is the eccentricity of the ellipse.

RealiseNothing · May 19, 2013

Re: HSC 2013 4U Marathon

asianese said:
d....x?

after looking at my solution again

Sy123 · May 19, 2013

Re: HSC 2013 4U Marathon

Trebla said:
I got a feeling I've asked this before, but will put it in anyway.

Prove that for any set of real numbers {x_i}

$\displaytsyle\sum_{k=1}^n x_i^2 \geq \dfrac{1}{n}\left(\displaystyle\sum_{k=1}^n x_i \right)^2$

Also, here is a Conics question for the purposes of variety on this thread:

The points P(a cos θ, b sin θ) and Q(a cos φ, b sin φ) lie on an ellipse. If PQ is a focal chord which passes through the positive focus show that the length of PQ is

$2a\left(1 - e \cos\dfrac{\theta + \psi}{2}\right)\left(1 + e \cos\dfrac{\theta + \psi}{2}\right)$

where e is the eccentricity of the ellipse.

Yep you have lol.

======

$$Let$ \ \ H_n = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$

$i) Prove that$

$H_n = \int_0^1 \frac{x^n - 1}{x - 1} \ dx$

$ii) Hence prove that$

$H_n = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} \binom{n}{k}$

seanieg89 · May 19, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$i) Prove that$ \ \ \int_{0}^{\frac{\pi}{2}} \sin^{2n-1} x \ dx = \frac{2 \cdot 4 \cdot 6 \dots (2n-2)}{1 \cdot 3 \cdot 5 \dots (2n-1)} \ \ \ \fbox{2}$

$ii) Hence or otherwise prove that$

$\sum_{n=1}^{\infty} \frac{1}{\binom{2n}{n}} = \frac{1}{3} + \frac{2\pi \sqrt{3}}{27} \ \ \ \ \fbox{6}$

I will provide a sketch of a proof. I encourage anyone who hasn't done this question to still try it though, it is good even when viewed as a purely algebraic exercise.

1. This is a standard integration by parts induction. We use I_n to denote this integral from now on.

2. Now the infinite series in ii) can be rewritten as the infinite series:

$\sum_{n\in\mathbb{N}}\frac{n}{2^{2n-1}}I_n$

3. For finite partial sums of this series, note that we can take the integral outside the sum. Use what we know about geometric series to find an expression for the thing inside this integral.

4. Split this integral into the sum of two integrals, one of which tends to zero as N -> infinity, and the other of which is independent of N. (Here N is the number of terms we are taking in our partial sum).

5. Flex our integration muscles. Its actually quite an easy integral, we have already done the hard work really.

6. QED bitches.

Note that this is an MX2 level proof, which could be shortened with more powerful tools. For example, the manual handling of the "error term" could be bypassed by using the monotone convergence theorem or some weaker result about the interchange of infinite summation and integration.

seanieg89 · May 19, 2013

Re: HSC 2013 4U Marathon

(Sorry for being so lazy, but I guess this thread is meant for current students. Hopefully my sketch will help someone actually doing their HSC to do the question!) If no-one uploads an actual answer I'll upload my handwritten solution later.

Sy123 · May 20, 2013

Re: HSC 2013 4U Marathon

$Prove that$

$\prod_{k=1}^{n} \cos \left(\frac{\pi(2k-1)}{4n} \right ) = \frac{\sqrt{2}}{2^n} \ \ \ \ \ \fbox{6}$

(please inform me if there is a mistake)

seanieg89 · May 20, 2013

Re: HSC 2013 4U Marathon

Trebla said:
I got a feeling I've asked this before, but will put it in anyway.

Prove that for any set of real numbers {x_i}

$\displaytsyle\sum_{k=1}^n x_i^2 \geq \dfrac{1}{n}\left(\displaystyle\sum_{k=1}^n x_i \right)^2$

$LHS=nx_i^2=RHS$

.

HSC 2013 MX2 Marathon (archive) (3 Viewers)

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