# Thread: HSC 2017 MX2 Marathon (archive)

1. ## Re: HSC 2017 4U Marathon

Are calculators allowed in the 4u test

2. ## Re: HSC 2017 4U Marathon

Originally Posted by si2136
Are calculators allowed in the 4u test
Yes.

3. ## Re: HSC 2017 4U Marathon

Prove De Moivre's Theorem by Mathematical Induction where |z|=1

4. ## Re: HSC 2017 4U Marathon

$\noindent Prove: \\\\ \left( x^{n-1} + \frac{1}{x^{n-1}} \right) + 2 \left( x^{n-2} + \frac{1}{x^{n-2}} \right) + \cdots + (n-1) \left( x + \frac{1}{x} \right) + n \equiv \frac{1}{x^{n-1}} \left(\frac{x^n -1}{x-1}\right)^2 \\\\ Hence, or otherwise, prove: \\\\ \cos{(n-1)\theta} + 2\cos{(n-2)\theta} + \cdots + (n-1)\cos{\theta} + \frac{n}{2} \equiv \frac{1}{2} \times \frac{1+\cos{n \theta}}{1+\cos{\theta}}$

5. ## Re: HSC 2017 4U Marathon

$\noindent Prove: \lim_{t \to 0} \left( 4e^{-t^4} -3e^{-t^3} - 2e^{-t^2} + e^{-t} \right)\log{t} = 0 \\\\ You may assume without proof that \lim_{t \to 0} t^a \log{t} = 0, \\\\ where a is any positive real number. \\\\ Using Integration by parts and the substitutions: x = t^4, y = t^3, z = t^2, \\\\ Prove: \int_0^\infty \frac{4e^{-t^4} - 3e^{-t^3} - 2e^{-t^2} + e^{-t}}{t} \text{d}t = 0$

6. ## Re: HSC 2017 4U Marathon

For the first part about the limit:

The best I could come up with was that (based on what my teacher said) is that all exponential functions dominate over all polynomial functions which dominate over all log functions. Therefore, based on the second line of the statement, because both g(t) (the function g denotes that left hand expression just before log(t)) and f(t) = t^a both tend to 0, and f(t) dominates over log(t), it must follow that the limit is true because g(t) also tends to 0 and takes precedence over the log function.

I'm still trying to practice on latex - I got the second part but I don't think it technically uses parts...

7. ## Re: HSC 2017 4U Marathon

Originally Posted by TaxiCab1729
For the first part about the limit:

The best I could come up with was that (based on what my teacher said) is that all exponential functions dominate over all polynomial functions which dominate over all log functions. Therefore, based on the second line of the statement, because both g(t) (the function g denotes that left hand expression just before log(t)) and f(t) = t^a both tend to 0, and f(t) dominates over log(t), it must follow that the limit is true because g(t) also tends to 0 and takes precedence over the log function.

I'm still trying to practice on latex - I got the second part but I don't think it technically uses parts...
You will need to flesh out your explanation more for it to be valid.

The idea is to use the inequalities:

$1+x \le e^x \le \frac{1}{1-x} \quad \forall x<1$

To bound the function behaviour near the origin and squeeze both sides.

8. ## Re: HSC 2017 4U Marathon

The idea is to use the inequalities:

$1+x \le e^x \le \frac{1}{1-x} \quad \forall x<1$

To bound the function behaviour near the origin and squeeze both sides.
Like, is anyone even gonna get this?

9. ## Re: HSC 2017 4U Marathon

Originally Posted by TaxiCab1729
Like, is anyone even gonna get this?
This is a marathon. They purposely throw in questions that can get overcomplicated. 100% of the time they are doable using just HSC techniques however if it were the HSC exam you'd be guided step by step

10. ## Re: HSC 2017 4U Marathon

$\noindent a) Prove \binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n-1}^2 + \binom{n}{n}^2 = \binom{2n}{n} \\\\ b) a,d are non-negative real numbers. a_0, a_1, \cdots , a_{n-1}, a_n, are non-negative real numbers in arithmetic progression. a_0 = a, and the common difference is d. Evaluate: \\\\ a_0 \binom{n}{0}^2 + a_1 \binom{n}{1}^2 + \cdots + a_{n-1} \binom{n}{n-1}^2 + a_n \binom{n}{n}^2$

11. ## Re: HSC 2017 4U Marathon

$\noindent a) Prove \binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n-1}^2 + \binom{n}{n}^2 = \binom{2n}{n} \\\\ b) a,d are non-negative real numbers. a_0, a_1, \cdots , a_{n-1}, a_n, are non-negative real numbers in arithmetic progression. a_0 = a, and the common difference is d. Evaluate: \\\\ a_0 \binom{n}{0}^2 + a_1 \binom{n}{1}^2 + \cdots + a_{n-1} \binom{n}{n-1}^2 + a_n \binom{n}{n}^2$
If anyone wants a hint, consider the expansion (1+x)^2n and what is is equal to for a), and for b), note that nCk=nC(n-k), and group together the like terms. Do consider the even and odd values of n separately for ease.

12. ## Re: HSC 2017 4U Marathon

Originally Posted by calamebe
If anyone wants a hint, consider the expansion (1+x)^2n and what is is equal to for a), and for b), note that nCk=nC(n-k), and group together the like terms. Do consider the even and odd values of n separately for ease.
There is no need to consider parity cases if you're at that level.

13. ## Re: HSC 2017 4U Marathon

There is no need to consider parity cases if you're at that level.
Eh, it works.

14. ## Re: HSC 2017 4U Marathon

\begin{align*}&\quad \sum_{k=0}^n(a+kd)\binom{n}{k} ^2\\ &= a\binom{2n}{n}+d \sum_{k=0}^nk\binom{n}{k} ^2\end{align*}

\text{Meh. Bit lost on how to tackle that without considering parity due to the existence of }\binom{n}{\frac{n}{2}} \\ \text{Odd:}\\ \begin{align*}&\sum_{k=0}^nk\binom{n}{k} ^2\\ &= \binom{n}{1}+2\binom{n}{2}+\dots+\frac{n-1}{2}\binom{n}{\frac{n-1}{2}}+\dots+(n-2)\binom{n}{n-2}+(n-1)\binom{n}{n-1}+n\binom{n}{n}\\ &= n\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k} \end{align*}\\ \text{but }\sum_{k=0}^{\frac{n-1}2}\binom{n}{k}=\frac{1}{2}\sum_{k=0}^n\binom{n}{ k}=2^n

And I probably did something wrong.

15. ## Re: HSC 2017 4U Marathon

Originally Posted by leehuan
\begin{align*}&\quad \sum_{k=0}^n(a+kd)\binom{n}{k} ^2\\ &= a\binom{2n}{n}+d \sum_{k=0}^nk\binom{n}{k} ^2\end{align*}

\text{Meh. Bit lost on how to tackle that without considering parity due to the existence of }\binom{n}{\frac{n}{2}} \\ \text{Odd:}\\ \begin{align*}&\sum_{k=0}^nk\binom{n}{k} ^2\\ &= \binom{n}{1}+2\binom{n}{2}+\dots+\frac{n-1}{2}\binom{n}{\frac{n-1}{2}}+\dots+(n-2)\binom{n}{n-2}+(n-1)\binom{n}{n-1}+n\binom{n}{n}\\ &= n\sum_{k=0}^{\frac{n-1}{2}}\binom{n}{k} \end{align*}\\ \text{but }\sum_{k=0}^{\frac{n-1}2}\binom{n}{k}=\frac{1}{2}\sum_{k=0}^n\binom{n}{ k}=2^n

And I probably did something wrong.
Yeah you left off the squares. Put the squares on, then use the result from a) and you should be fine.

EDIT: Though I don't see a need to break the terms up as you did in the second line. You can do it easily without breaking it up.

16. ## Re: HSC 2017 4U Marathon

Originally Posted by calamebe
Yeah you left off the squares. Put the squares on, then use the result from a) and you should be fine.

EDIT: Though I don't see a need to break the terms up as you did in the second line. You can do it easily without breaking it up.
Was wondering where the squares vanished to.

There wasn't a need for it; I did it purely for my visual purposes.

17. ## Re: HSC 2017 4U Marathon

You really don't need to split based upon parity. It only makes things worse.

You haven't considered

$\binom{n}{k} = \binom{n}{n-k}$

enough.

Side note your conclusion is wrong.

18. ## Re: HSC 2017 4U Marathon

Here's a nice anecdote.

"One day Gauss' teacher asked his class to add together all the numbers from 1 to 100, assuming that this task would occupy them for quite a while. He was shocked when young Gauss, after a few seconds thought, wrote down the answer 5050. The teacher couldn't understand how his pupil had calculated the sum so quickly in his head, but the eight year old Gauss pointed out that the problem was actually quite simple.

He had added the numbers in pairs - the first and the last, the second and the second to last and so on, observing that 1+100=101, 2+99=101, 3+98=101, ...so the total would be 50 lots of 101, which is 5050"

19. ## Re: HSC 2017 4U Marathon

Damn it...I was doing that

20. ## Re: HSC 2017 4U Marathon

Binomial Question.pdf

I think I got it, but it looks dodgy...

21. ## Re: HSC 2017 4U Marathon

Originally Posted by TaxiCab1729
Binomial Question.pdf

I think I got it, but it looks dodgy...
For odd n, I don't think your S_2 is an integer. I think you don't need that subtraction by 1.

Also note that you can evaluate the sums C(n,k)^2 and kC(n,k)^2 combinatorially. (This takes less effort than other methods of computation imo.)

These sums amount to:

A: The number of ways of choosing a team of n people out of 2n people, n of whom are in year 12 and n of whom are in year 11.

This can be done in C(2n,n) ways.

B: Same as above, but also selecting one of the year 12 students to be team captain.

This can be done in nC(2n,n)/2 ways. (The number of ways of choosing a team and specifying a captain is clearly nC(2n,n). By symmetry, this captain will be in year 12 exactly half of the time.)

22. ## Re: HSC 2017 4U Marathon

Note that the above yields an end answer of something like

$\left(a+\frac{nd}{2}\right)\binom{2n}{n}.$

Busy prepping for a meeting right now but paradoxica can confirm if that is what he was looking for.

23. ## Re: HSC 2017 4U Marathon

Originally Posted by seanieg89
Note that the above yields an end answer of something like

$\left(a+\frac{nd}{2}\right)\binom{2n}{n}.$

Busy prepping for a meeting right now but paradoxica can confirm if that is what he was looking for.
Yeah that's what I got too.

24. ## Re: HSC 2017 4U Marathon

Yeah...I realized my mistake. My bad...

25. ## Re: HSC 2017 4U Marathon

For z^5-1=0, the roots are 1, w^2-w^4, and
z^5-1=(z-1)(z-cis2pi/5)(z-cis-2pi/5)(z-cis4pi/5)(z-cis-4pi/5) How do you get back to the LHS from expanding the RHS?

Page 2 of 5 First 1234 ... Last

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•