HSC 2015 MX1 Marathon (archive) (1 Viewer)

rand_althor · Aug 27, 2015

Re: HSC 2015 3U Marathon

sharoooooo said:
help me pls with this question pls pls

Q. Consider the digits 9,8,7,6,5,4,3,2,1 and 0. Find how many five-digit numbers are possible if the digits are to be in:
a) i) ascending order
a) ii) descending order
b) Why do these 2 questions involve unordered selections?

Answer:
a) i) 126
a) ii) 252

pls halp i dont understand

b) For each, you pick a group of numbers where order does not matter. Once you have that group, there is only one way of arranging them in either ascending or descending order.

davidgoes4wce · Aug 27, 2015

Re: HSC 2015 3U Marathon

Anyone I'm tired its been another fun day of maths. Good night everyone.

sharoooooo · Aug 27, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
a)
Pick any 5 numbers 'Once' from 1 to 9 (can't choose 0) . 9C5=126

b)
Pick any 5 Numbers 'Once' from 1 to 10 (not ordering. 10C5=252

rand_althor said:
b) For each, you pick a group of numbers where order does not matter. Once you have that group, there is only one way of arranging them in either ascending or descending order.

thanks

davidgoes4wce · Aug 31, 2015

Re: HSC 2015 3U Marathon

Decided I'm going to make a donation and upgrade to 'Premier Resources' tomorrow. This site has helped make me a better maths student. When you study HSC you need every advantage you can get and this site is definitely one of them. Thanks mods and people that contribute to these threads

davidgoes4wce · Aug 31, 2015

Re: HSC 2015 3U Marathon

Was having a deep think about this question this evening and still didn't quite get it.

I was thinking of 'dominant terms' but my value does not match the back of the book.

Drsoccerball · Aug 31, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Was having a deep think about this question this evening and still didn't quite get it.

I was thinking of 'dominant terms' but my value does not match the back of the book.

its 0

davidgoes4wce · Aug 31, 2015

Re: HSC 2015 3U Marathon

Really? I was thinking that as well but the solution has 1/4 as the answer.

rand_althor · Aug 31, 2015

Re: HSC 2015 3U Marathon

$$There is an induction proof which shows that $1^3+2^3+3^3+4^3+...+n^3=\frac{n^2(n+1)^2}{4}$. If you use this, you can write the limit as:$

$\begin{align*}\lim_{n\to\infty} \frac{1^3+2^3+3^3+4^3+...+n^3}{n^4} &= \lim_{n\to\infty} \frac{\left (\frac{n^2(n+1)^2}{4} \right )}{n^4} \\ &= \lim_{n\to\infty} \frac{n^2(n^2+2n+1)}{4n^4} \\ &= \lim_{n\to\infty} \frac{n^2+2n+1}{4n^2} \\&= \lim_{n\to\infty} \frac{1}{4}+\frac{1}{2n}+\frac{1}{4n^2} \\&=\frac{1}{4}\end{align*}$

However, if you are not given that information, I don't know how to do it.

VBN2470 · Aug 31, 2015

Re: HSC 2015 3U Marathon

The answer is 1/4. The sum $1^3+2^3+...+n^3$ can be simplified to $\frac{n^2(n+1)^2}{4}$ , which you can substitute into the expression and simplify accordingly.

InteGrand · Aug 31, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$There is an induction proof which shows that $1^3+2^3+3^3+4^3+...+n^3=\frac{n^2(n+1)^2}{4}$. If you use this, you can write the limit as:$

$\begin{align*}\lim_{n\to\infty} \frac{1^3+2^3+3^3+4^3+...n^3}{n^4} &= \lim_{n\to\infty} \frac{\left (\frac{n^2(n+1)^2}{4} \right )}{n^4} \\&= \lim_{n\to\infty} \frac{n^2+2n+1}{4n^2} \\&= \lim_{n\to\infty} \frac{1}{4}+\frac{1}{2n}+\frac{1}{4n^2} \\&=\frac{1}{4}\end{align*}$

However, if you are not given that information, I don't know how to do it.

$$One way to remember that formula for the sum of cubes is that it happens to be the square of: $1 + 2 + 3 + \cdots+ n$ (which is easily remembered to be $\frac{n(n+1)}{2}$ (or found using AP sum formula)).$

Drsoccerball · Aug 31, 2015

Re: HSC 2015 3U Marathon

How do you derive the formula of the sum of cubes ?

VBN2470 · Aug 31, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
How do you derive the formula of the sum of cubes ?

Here's is one method: http://mathschallenge.net/library/number/sum_of_cubes

InteGrand · Aug 31, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
How do you derive the formula of the sum of cubes ?

https://en.wikipedia.org/wiki/Squared_triangular_number

InteGrand · Aug 31, 2015

Re: HSC 2015 3U Marathon

VBN2470 said:
Here's is one method: http://mathschallenge.net/library/number/sum_of_cubes

Lol I just saw that site myself too. And it's a similar method for deriving the formula for the sum of first n squares.

davidgoes4wce · Aug 31, 2015

Re: HSC 2015 3U Marathon

That was the original question :

InteGrand · Aug 31, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
That was the original question :

$LHS=\frac{2^2 -1}{2^2}\cdot \frac{3^2 -1}{3^2}\cdot \frac{4^2 -1}{4^2}\cdot \cdots \cdot \frac{n^2 -1}{n^2}$

$$=\frac{(2+1)(2-1)}{2^2}\cdot \frac{(3+1)(3-1)}{3^2}\cdot \frac{(4+1)(4-1)}{4^2}\cdot \cdots \cdot \frac{(n+1)(n-1)}{n^2}$ (difference of two squares)$

$=\frac{3\times 1}{2^2}\cdot \frac{4\times 2}{3^2}\cdot \frac{5\times 3}{4^2}\cdot \cdots \cdot \frac{(n+1)(n-1)}{n^2}$

$=\frac{\left(3\times 4 \times 5 \times \cdots \times (n+1) \right)\left(1\times 2 \times 3 \times \cdots \times (n-1) \right)}{\left(2\times 3 \times \cdots \times n \right)\cdot \left(2\times 3 \times \cdots \times n \right)}$

$$=\frac{n+1}{2n}$ (by cancellation of terms from the numerator and denominator)$

$$=RHS$.$

davidgoes4wce · Aug 31, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$There is an induction proof which shows that $1^3+2^3+3^3+4^3+...+n^3=\frac{n^2(n+1)^2}{4}$. If you use this, you can write the limit as:$

$\begin{align*}\lim_{n\to\infty} \frac{1^3+2^3+3^3+4^3+...+n^3}{n^4} &= \lim_{n\to\infty} \frac{\left (\frac{n^2(n+1)^2}{4} \right )}{n^4} \\ &= \lim_{n\to\infty} \frac{n^2(n^2+2n+1)}{4n^4} \\ &= \lim_{n\to\infty} \frac{n^2+2n+1}{4n^2} \\&= \lim_{n\to\infty} \frac{1}{4}+\frac{1}{2n}+\frac{1}{4n^2} \\&=\frac{1}{4}\end{align*}$

However, if you are not given that information, I don't know how to do it.

It's interesting that you mentioned that , the previous question before, was for that specific proof. But if your saying its good to remember that I will try to remember it in future.

InteGrand · Aug 31, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
It's interesting that you mentioned that , the previous question before, was for that specific proof. But if your saying its good to remember that I will try to remember it in future.

$For the HSC, there isn't much point in memorising that result. If it's useful for a question, there will most likely be a part where they get you to prove it first (most likely by induction (perhaps ``or otherwise'' also)).$

Jessica Norton · Sep 1, 2015

Re: HSC 2015 3U Marathon

Answer please ASAP
View attachment 32350

davidgoes4wce · Sep 1, 2015

Re: HSC 2015 3U Marathon

Part
i)

P (Andy wins 2 sets, Novak wins 2 sets)= 4C2 x (2/3)^2 x (1/3)^2= 24/81

with regards to part (ii)

I have uploaded my screenshot. Feel free if anyone sees any error in my work to correct me

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