HSC 2013 MX2 Marathon (archive) (3 Viewers)

Marc26 · Dec 30, 2012

Re: HSC 2013 4U Marathon

Can someone help me here? I'm struggling to understand

HSC 2000 - Q8

RealiseNothing · Dec 30, 2012

Re: HSC 2013 4U Marathon

Marc26 said:
Can someone help me here? I'm struggling to understand

HSC 2000 - Q8

View attachment 27306

Sum each one individually:

$z = \frac{z(z-1)}{z-1}$

$z + z^2=\frac{z(z^2-1)}{z-1}$

$z + z^2 + z^3 = \frac{z(z^3-1)}{z-1}$

...

$z + z^2 + z^3 + ... + z^n = \frac{z(z^n-1)}{z-1}$

Add these all up now since they all have a common denominator:

$\frac{z(z-1) + z(z^2-1) + z(z^3-1) + ... + z(z^n-1)}{z-1}$

$\frac{z(z-1+z^2-1+z^3-1+...z^n-1)}{z-1}$

$\frac{z(z+z^2+z^3+...+z^n - n)}{z-1}$

$\frac{z}{z-1} \times \frac{z(z^n-1)}{z-1} - \frac{nz}{z-1}$

$\frac{nz}{1-z} + \frac{z^2(z^n-1)}{(z-1)^2}$

$\frac{nz}{1-z} - \frac{z^2}{(1-z^2)}(1-z^n)$

Demento1 · Dec 30, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Question Pell - 15 marks$

$Define the Pell numbers by the recursive relation$

$P_n=\begin{cases} 0 \ \ \ \ \ $if$ \ \ n = 0 \\ 1 \ \ \ \ \ $if$ \ \ n = 1 \\ 2P_{n-1}+P_{n-2} \ \ \ \ \ n \geq 2 \end{cases}$

$$i) Find the first 5 Pell numbers$ \ \ \ \ \ \fbox{1}$

$$ii) Prove using Mathematical Induction that$ \\ \\ P_{n+1}P_{n-1}-P_n^2=(-1)^n \ \ \ \ \ $for$ \ \ n \geq 1 \ \ \ \ \ \fbox{3}$

$$iii) Hence prove that $ \ \ \ x=P_{2n-1}+P_{2n} \ \ $and$ \ \ y= P_{2n} \ \ \ \ $satisfy the equation for all positive integers n$ \\ \\ x^2-2y^2=1 \ \ \ \fbox{3}$

$$The kth triangular number can be found by$ \ \ \ t_k = \frac{k(k+1)}{2} \ \ \ $and the mth square number can be found by$ \ \ \ s_m = m^2$
$Let $ \ \ N_p \ \ $denote the pth number that is both triangular and square$
$$Hence there exists integers, k and m such that$ \ \ m^2=\frac{k(k+1)}{2} = N_p$

$$iv) Show that$ \ \ \ x=2k+1 \ \ \ y=2m \\ $satisfy the equation$ \ \ \ x^2-2y^2=1 \ \ \ \fbox{3}$

$$v) Hence using part (iii) and (iv), show that N_p = \left ( \frac{P_{2p}}{2} \right ) ^2 \ \ \ \fbox{2}$

$$vi) Considering the equality$ \ \ \ \ \lim_{j \to \infty} \frac{P_{j+1}}{P_j} = \lim_{j \to \infty} \frac{P_{j}}{P_{j-1}}$

$$Find$ \ \ \lim_{p\to \infty} \frac{N_{p+1}}{N_p} \ \ \ \ \fbox{3}$

For Question Pell:

i) 1, 1, 3, 7, 17

Sorry, not familiar with induction...

Sy123 · Dec 30, 2012

Re: HSC 2013 4U Marathon

Demento1 said:
For Question Pell:

i) 1, 1, 3, 7, 17

Sorry, not familiar with induction...

Your first value is wrong -> which has made the whole sequence wrong.

Demento1 · Dec 30, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Your first value is wrong -> which has made the whole sequence wrong.

Wait, we actually count the zero as the first one?

If so, then yeah it should be: 0, 1, 2, 5, 12

Sy123 · Dec 30, 2012

Re: HSC 2013 4U Marathon

Demento1 said:
Wait, we actually count the zero as the first one?

If so, then yeah it should be: 0, 1, 2, 5, 12

Correct, but those are the easy marks :s

For part iii, use the result in part ii.

GoldyOrNugget · Dec 30, 2012

Re: HSC 2013 4U Marathon

Doing them on paper, won't TeX up every step.

i) 0, 1, 2, 5, 12

ii) $R.T.P. $P_{k+2}P_k - P_{k+1}^2 = (-1)^{k+1}$, having assumed that $P_{k+1}P_{k-1} - P_k^2 = (-1)^{k}$. Express everything in terms of $P_k, P_{k-1}$ by using the recurrence, then you get -RHS = LHS which is what you want.$

iii) Subbing in x,y and expanding, you get the RHS from part ii. Now because you're using n=2k and k is integral, (-1)^(2k) = 1 as required.

iv) Sub in the given expressions for x and y, then sub in m² = k(k+1)/2. It cancels to equal 1.

v) The desired equation can be reached by equating the two expressions given for y to get an expression for m, then squaring this to get N_p. However, I'm reluctant to do this, because the fact that a and b both satisfy an equation does not mean that a = b. I think my working would be sufficient in an exam, but I'd need to justify this step to myself properly.

vi) By manually computing a few more Pell numbers, It's easy to convince oneself that the ratio between consecutive terms converges to 1 + sqrt(2) (not very mathematical, hehe). Now expanding the limit we're meant to find, we get (P_2p+2 / P_2p)². The nice limit equalit means that this ratio by itself is (1+sqrt(2))^2 -- squaring this again gives (1+sqrt(2))^4 which is the answer.

Sy123 · Dec 30, 2012

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Doing them on paper, won't TeX up every step.

i) 0, 1, 2, 5, 12

ii) $R.T.P. $P_{k+2}P_k - P_{k+1}^2 = (-1)^{k+1}$, having assumed that $P_{k+1}P_{k-1} - P_k^2 = (-1)^{k}$. Express everything in terms of $P_k, P_{k-1}$ by using the recurrence, then you get -RHS = LHS which is what you want.$

iii) Subbing in x,y and expanding, you get the RHS from part ii. Now because you're using n=2k and k is integral, (-1)^(2k) = 1 as required.

iv) Sub in the given expressions for x and y, then sub in m² = k(k+1)/2. It cancels to equal 1.

v) The desired equation can be reached by equating the two expressions given for y to get an expression for m, then squaring this to get N_p. However, I'm reluctant to do this, because the fact that a and b both satisfy an equation does not mean that a = b. I think my working would be sufficient in an exam, but I'd need to justify this step to myself properly.

vi) By manually computing a few more Pell numbers, It's easy to convince oneself that the ratio between consecutive terms converges to 1 + sqrt(2) (not very mathematical, hehe). Now expanding the limit we're meant to find, we get (P_2p+2 / P_2p)². The nice limit equalit means that this ratio by itself is (1+sqrt(2))^2 -- squaring this again gives (1+sqrt(2))^4 which is the answer.

Correct - Nice work. I am working on a Complex Numbers question at the moment, should be good.

GoldyOrNugget · Dec 30, 2012

Re: HSC 2013 4U Marathon

Here's a tricky little perms & combs for y'all.

$i) There are 5 varieties of fruit available (strawberry, apple, pear, grape, melon). I want to buy at least 1 fruit and at most 12 fruit. In how many ways can I do this? (e.g. one such purchase could be 2 strawberries, an apple, 3 grapes and 5 melons)$

$ii) I now do not distinguish between combinations where proportions are identical. e.g. purchasing an apple and 3 grapes is the same as purchasing 2 apples and 6 grapes, but is different to purchasing 2 apples, 6 grapes and a melon. Again, I want to buy at least 1 and at most 12 fruit. In how many ways can I do this?$

Sy123 · Dec 30, 2012

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
Here's a tricky little perms & combs for y'all.

$i) There are 5 varieties of fruit available (strawberry, apple, pear, grape, melon). I want to buy at least 1 fruit and at most 12 fruit. In how many ways can I do this? (e.g. one such purchase could be 2 strawberries, an apple, 3 grapes and 5 melons)$

$ii) I now do not distinguish between combinations where proportions are identical. e.g. purchasing an apple and 3 grapes is the same as purchasing 2 apples and 6 grapes, but is different to purchasing 2 apples, 6 grapes and a melon. Again, I want to buy at least 1 and at most 12 fruit. In how many ways can I do this?$

For the first one I got 425

RealiseNothing · Dec 30, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
For the first one I got 425

Surely it's more than that though.

RealiseNothing · Dec 30, 2012

Re: HSC 2013 4U Marathon

Using triangular numbers I ended up with:

$5 + t_5 + \sum_{k=1}^{5}t_k + \sum_{n=1}^{5} \sum_{k=1}^{n}t_k + ...$

But this was way too tedious to evaluate.

Sy123 · Dec 31, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Using triangular numbers I ended up with:

$5 + t_5 + \sum_{k=1}^{5}t_k + \sum_{n=1}^{5} \sum_{k=1}^{n}t_k + ...$

But this was way too tedious to evaluate.

The figurate numbers give the triangular numbers, sum of tri numbers (tetra), sum of that, and so on:
The cool thing is that they can all be expressed in the form of a binomial co-efficient:

http://en.wikipedia.org/wiki/Figurate_number

I had another look at it, the counting logic that I had was wrong, but the overall concept I had I think is right.

Let case k be the case where I pick k fruit:

Case 1: Pick 1 fruit - 5 ways = 5C1

Case 2: Pick 2 fruit - (Case 1) + 10

Now the reason why Case 1 is part of this sum, is that when we pick 2 fruit, we can pick 'doubles'
i.e. 2 strawberries, 2 oranges etc.

Which is why there is the same amount of Case 1 permutations, this concept was the basis of my answer.
The other 10 came from 5C2

Case 3: Pick 3 - (Case 1) + (Case 2) + 5C3

We have Case 1 there because we can pick triple fruit, we can pick Case 2 becasue we just add on one fruit to each permutation from Case 2 (not sure how to explain this properly without a diagram), then 5C3

Case 4: Pick 4 - (Case 1) + (Case 2) + (Case 3) + 5C4

Case 5: Pick 5 - $\sum_{k=1}^{4} $Case$ \ k$ + 5C5

And here is where things change a bit:

Case 6: Pick 6 - (Sum Case 1 to Case 5) + 5C1

again we are simply picking 1 more fruit, so we can pick this fruit from any of the 5 types.

Case 7: Pick 6 - (Sum Case 1 to Case 6) + 5C2

...

Case 10: Pick 10 - (Sum Case 1 to Case 10) + 5C5

Then the cycle repeats

Case 12: (Sum Case 1 to Case 11) + 5C2

==========

So the figure must be huge.
If you are bothered to check my explanations, is there anything wrong with it?

GoldyOrNugget · Dec 31, 2012

Re: HSC 2013 4U Marathon

The answer is 4 digits long. I've verified it. It shouldn't be tedious to compute -- I did it quickly using only the operations available on a standard BOS-approved calculator.

Part ii) perhaps requires slightly more logical reasoning than you'd get in an HSC paper, but part i) is about the same difficulty as the perms+combs HSC questions of past years.

seanieg89 · Dec 31, 2012

Re: HSC 2013 4U Marathon

$i) For each positive integer $k$, the number of ways of choosing $k$ fruits out of our $5$ types is \\ $\frac{(k+4)!}{k!4!}$\\ by counting strings consisting of $k$ dots and $4$ dividing slashes.\\Hence the number of ways of choosing some nonzero quantity of fruit at most some positive integer $n$ is given by:\\ \\ $\frac{1}{4!}\sum_{k=1}^n \frac{(k+4)!}{k!}=\frac{1}{4!}\sum_{k=1}^n \frac{1}{5}(\frac{(k+5)!}{k!}-\frac{(k+4)!}{(k-1)!})\\=\frac{1}{5!}\left(\frac{(n+5)!}{n!}-5!\right).$\\by telescoping. Chuck in $n=12$ to complete.$

Sy123 · Dec 31, 2012

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
The answer is 4 digits long. I've verified it. It shouldn't be tedious to compute -- I did it quickly using only the operations available on a standard BOS-approved calculator.

Part ii) perhaps requires slightly more logical reasoning than you'd get in an HSC paper, but part i) is about the same difficulty as the perms+combs HSC questions of past years.

Is my reasoning correct?
I need to get better at perms and combs

GoldyOrNugget · Dec 31, 2012

Re: HSC 2013 4U Marathon

seanie, Correct!

I used the same dots/slashes reasoning, but phrased it as 'choosing slash positions between dots'. For some fixed number of fruit to purchase 'n', this is given by ^n+5-1C_5-1. Once you get into the rhythm of button-mashing, it only takes a few seconds to type in 5C4+6C4+7C4+8C4+9C4+10C4+11C4+12C4+13C4+14C4+15C4+16C4. As Sy pointed out, these individual choose terms are figurate numbers -- specifically, the next dimension up from tetrahedral numbers. You can probably do some maths based on that to avoid all the typing.

I'll be interested to see approaches to part ii)...

GoldyOrNugget · Dec 31, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Is my reasoning correct?
I need to get better at perms and combs

It looks like you have some good ideas, but it needs refining. For your case 3, for example, the correct answer is 35, but your method gives 30. I think it's because of this: "we can pick Case 2 becasue we just add on one fruit to each permutation from Case 2". But case 2 includes case 1, so when you add one fruit to one of case 2's permutations that it borrowed from case 1, you'll now have one fruit picked 3 times, which will be double-counted when you separately add on case 1. Also, you'd need to account for which one of the fruit in the pair that you add one to -- but again, you can't just multiply by 2, because that screws up the permutations case 2 borrowed from case 1 where two of the same fruit were picked.

Sy123 · Dec 31, 2012

Re: HSC 2013 4U Marathon

Let us establish the fact that if we pick an odd number of fruits, we cannot retain proportionality. Hence we keep all the odd numbered fruits arrangements (odd numbers above 6, I will get to this later)

Now, for even number of fruits picked, for example say we pick 3 fruits, the total number of ways to pick 3 fruits is 35.
Now consider picking 6 fruits (we doubled it), now we can definitely say that proportionality of the ways to pick 3 fruits remain.

Hence the new total permuations for 6 fruits chosen is
(Arrangements for 6 normally) - 35

And when we add them up, the +35 from the arrangements for 3 will cancel out with the -35.

But, for instance, we can't double 7 since we get 14 > 12

Hence we add up the arrangements of 7, 8, 9, 10, 11, 12

11C4 + 12C4 + 13C4 + 14C4 + 15C4 + 16C4 = 5726

(probably wrong, but I reckon the logic is right)

GoldyOrNugget · Dec 31, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Let us establish the fact that if we pick an odd number of fruits, we cannot retain proportionality.

Sounds good

Sy123 said:
Now consider picking 6 fruits (we doubled it), now we can definitely say that proportionality of the ways to pick 3 fruits remain.

Hence the new total permuations for 6 fruits chosen is
(Arrangements for 6 normally) - 35

Hm ok... but what about the arrangement (3 grapes, 3 apples)? This is an arrangement for 6 fruits, and it's not proportional to any arrangement of 3 fruits , but yet it isn't legal...

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HSC 2013 MX2 Marathon (archive) (3 Viewers)

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