HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Sep 27, 2013

Re: HSC 2013 4U Marathon

harryharper said:
Is it $p+q\le1$ ?

$x^2+y^2\ge2xy$

$x^2(\frac{1-p}{q})+y^2(\frac{1-p}{q})\ge2xy$

$x^2(1-p)p+y^2(1-q)q\ge2pqxy$

$px^2+qy^2\ge p^2x^2+q^2y^2+2pqxy$

$px^2+qy^2\ge (px+qy)^2$

Yep, well done

---

$$i)$ \ \ \sum_{k=1}^{\infty} \frac{1}{k(k+2)} \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{k+2} \right )$

$$ii)$ \ \ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{mn^2 + 2mn + m^2n}$

harryharper · Sep 27, 2013

Re: HSC 2013 4U Marathon

Part i: first step change it to 11/8 + 0.5*Sum from r=3 to infinite of (2r-1)/(r^3-r). This comes from 'changing the perspective' from summing sums of 1/k terms to sums of 1/((k+2)k) terms.
Then use partial fractions and a couple of telescopes to get 11/8+5/12-1/24=7/4 as the answer.

Sy123 · Sep 27, 2013

Re: HSC 2013 4U Marathon

harryharper said:
Part i: first step change it to 11/8 + Sum from r=3 to infinite of (2r-1)/(r^3-r)
Then use partial fractions and a couple of telescopes to get 11/8+5/12-1/24=7/4 as the answer.

How did you get (2r-1)/(r^3-r), that is the correct answer though, I split the 1/k(k+2) to 1/2 ( 1/k - 1/(k+2)), then manipulated and grouped certain terms to telescope twice.

harryharper · Sep 27, 2013

Re: HSC 2013 4U Marathon

Intermediate steps included:

11/8 + Sum r=3 to infinite: 1/(r+1)*(1/((r-1)(r+1))+1/(r(r+2))+1/((r+1)(r+3))+....)

Which equals

11/8 + 0.5*Sum r=3 to infinite: 1/(r+1)*(1/(r-1)+1/r)

Sorry no Latex - still working on my skills in that department.

Sy123 · Sep 28, 2013

Re: HSC 2013 4U Marathon

harryharper said:
Intermediate steps included:

11/8 + Sum r=3 to infinite: 1/(r+1)*(1/((r-1)(r+1))+1/(r(r+2))+1/((r+1)(r+3))+....)

Which equals

11/8 + 0.5*Sum r=3 to infinite: 1/(r+1)*(1/(r-1)+1/r)

Sorry no Latex - still working on my skills in that department.

Ah yep ok that is quite similar to what I did.

Sy123 · Sep 28, 2013

Re: HSC 2013 4U Marathon

$$Prove for real$ \ \ \{ a_k \} \ \ $and$ \ \ \{b_k \}$

$\sum_{k=1}^{n} \sqrt{a_k^2+ b_k^2} \geq \sqrt{ \left(\sum_{k=1}^n a_k \right)^2 + \left(\sum_{k=1}^n b_k \right )^2}$

harryharper · Sep 28, 2013

Re: HSC 2013 4U Marathon

This one's the triangle inequality right? Just need a little bit of induction to extend it.

Sy123 · Sep 28, 2013

Re: HSC 2013 4U Marathon

harryharper said:
This one's the triangle inequality right? Just need a little bit of induction to extend it.

I don't think so

harryharper · Sep 28, 2013

Re: HSC 2013 4U Marathon

$$z_{k}=a_{k}+ib_{k}$

$\text{so the inequality says}$

$\sum_{k=1}^{n}|z_{k}|\ge\left|\sum_{k=1}^{n}z_{k}\right|$

$n=2\text{ is the triangle inequality and the induction is straightforward}$

$\left|z_{m+1}|+\sum_{k=1}^{m}|z_{k}|\ge|z_{m+1}|+|\sum_{k=1}^{m}z_{k}\right|\ge\left|z_{m+1}+\sum_{k=1}^{m}z_{k}\right|=\left|\sum_{k=1}^{m+1}z_{k}|$

Sy123 · Sep 28, 2013

Re: HSC 2013 4U Marathon

harryharper said:
$$z_{k}=a_{k}+ib_{k}$

$\text{so the inequality says}$

$\sum_{k=1}^{n}|z_{k}|\ge\left|\sum_{k=1}^{n}z_{k}\right|$

$n=2\text{ is the triangle inequality and the induction is straightforward}$

$\left|z_{m+1}|+\sum_{k=1}^{m}|z_{k}|\ge|z_{m+1}|+|\sum_{k=1}^{m}z_{k}\right|\ge\left|z_{m+1}+\sum_{k=1}^{m}z_{k}\right|=\left|\sum_{k=1}^{m+1}z_{k}|$

Ohhh I see what you mean now, nice work.

I squared both sides and cancelled out the a_k^2 and b_k^2 terms, and then by grouping the terms

proving

$\sqrt{(a_k^2 + b_k^2)(a_m^2 + b_m^2)} \geq a_k a_m + b_k b_m$

Which is done by squaring both sides and expanding, to yield

$a_k^2b_m^2 + a_m^2b_k^2 \geq 2a_ka_m b_k b_m$

Which is just the AM-GM inequality.

Your method is more elegant though lol

----

$Prove that there are no positive integers$ \ \ (a,b) \ \ $that satisfy$

$b^2 + b + 1 = a^2$

harryharper · Sep 28, 2013

Re: HSC 2013 4U Marathon

$b^2<b^2+b+1<b^2+2b+1=(b+1)^2$

No squares between consecutive squares.

harryharper · Sep 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$ii)$ \ \ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{mn^2 + 2mn + m^2n}$

$\ \ \sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\\frac{1}{m}\frac{1}{m+n+2}$

$\ \ \sum_{n=1}^{\infty}\frac{1}{n(n+2)}\sum_{m=1}^{ \infty }\frac{1}{m}-\frac{1}{m+n+2}$

$\ \ \sum_{n=1}^{\infty}\frac{1}{n(n+2)}\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n+2})$

$which is i)$

harryharper · Sep 28, 2013

Re: HSC 2013 4U Marathon

There are M marbles in a jar, R of which are red. I draw out N marbles without replacing any.

i) What is the probability of me drawing out K red marbles?

ii) What is the most likely number of red marbles I will draw out?

iii) What is the average number of red marbles I will draw out?

Sy123 · Sep 28, 2013

Re: HSC 2013 4U Marathon

harryharper said:
There are M marbles in a jar, R of which are red. I draw out N marbles without replacing any.

i) What is the probability of me drawing out K red marbles?

ii) What is the most likely number of red marbles I will draw out?

iii) What is the average number of red marbles I will draw out?

Here is an attempt:

$$i)$ \ \ \ \frac{C(r,k) \times C(m-r,n-k)}{C(m,n)}$

$$ii) \ \ $let$ \ \ T_{k} = \binom{r}{k} \binom{m-r}{n-k}$

$\frac{T_{k+1}}{T_{k}} > 1$

$\Rightarrow \ \ \frac{nr-m+r+n-1}{m+2} > k$

$Therefore the$ \ \ k \ \ $that maximizes probability, thus the most likely is$

$k = \left \lfloor \frac{nr - m + r + n- 1}{m+2} \right \rfloor$

$$Where$ \ \ \lfloor x \rfloor \ \ $is the greatest integer less than or equal to$ \ \ x$

harryharper · Sep 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Here is an attempt:

$$i)$ \ \ \ \frac{C(r,k) \times C(m-r,n-k)}{C(m,n)}$

$$ii) \ \ $let$ \ \ T_{k} = \binom{r}{k} \binom{m-r}{n-k}$

$\frac{T_{k+1}}{T_{k}} > 1$

$\Rightarrow \ \ \frac{nr-m+r+n-1}{m+2} > k$

$Therefore the$ \ \ k \ \ $that maximizes probability, thus the most likely is$

$k = \left \lfloor \frac{nr - m + r + n- 1}{m+2} \right \rfloor$

$$Where$ \ \ \lfloor x \rfloor \ \ $is the greatest integer less than or equal to$ \ \ x$

i) is correct.

ii) is the correct approach, and very nearly the correct answer - I suspect some minor error somewhere along the line. The numerator should be (n+1)(r+1)=nr+n+r+1

Sy123 · Sep 28, 2013

Re: HSC 2013 4U Marathon

harryharper said:
i) is correct.

ii) is the correct approach, and very nearly the correct answer - I suspect some minor error somewhere along the line. The numerator should be (n+1)(r+1)=nr+n+r+1

Ahh ok, I'm not too sure about how to interpret 'average' though..

If I did:

$P(k) = \frac{C(r,k) \times C(m-r,n-k)}{C(m,n)}$

And took the average probability:

$A = \frac{1}{n} \sum_{k=1}^r P(k)$

And then, if:

$P(m) < A < P(m+1)$ (so its bounded by two consecutive probabilities)

Then the one its closest to, (say its closest to P(m)), that means m is the average number of balls????

harryharper · Sep 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Ahh ok, I'm not too sure about how to interpret 'average' though..

If I did:

$P(k) = \frac{C(r,k) \times C(m-r,n-k)}{C(m,n)}$

And took the average probability:

$A = \frac{1}{n} \sum_{k=1}^r P(k)$

And then, if:

$P(m) < A < P(m+1)$ (so its bounded by two consecutive probabilities)

Then the one its closest to, (say its closest to P(m)), that means m is the average number of balls????

$Ok, so by the average number of balls, I'm referring to the expected number of balls: the weighted average of all possible values.$

$That is,$

$$E(x)=\sum_{k=0}^r kP(k)$

$\text{Interesting to think about what m would be in your formulation.}$

$\text{Clearly } A=\frac{1}{n} $, since$ \sum_{k=0}^r P(k)=1$

$Hmm...not so sure now if finding this m would be particularly meaningful. My E(x) clarification still stands.$

Sy123 · Sep 28, 2013

Re: HSC 2013 4U Marathon

harryharper said:
$Ok, so by the average number of balls, I'm referring to the expected number of balls: the weighted average of all possible values.$

$That is,$

$$E(x)=\sum_{k=0}^r kP(k)$

$\text{Interesting to think about what m would be in your formulation.}$

$\text{Clearly } A=\frac{1}{n} $, since$ \sum_{k=0}^r P(k)=1$

$Hmm...not so sure now if finding this m would be particularly meaningful. My E(x) clarification still stands.$

Oh ok

As for calculation of E(x)

$Finding the sum$

$\sum_{k=0}^r k \binom{r}{k} \binom{m-r}{n-k} k$

Consider the identity:

$r(1+x)^{r-1} (1+x)^{m-r} = r (1+x)^{m-1}$

Equate co-efficient of $x^{n-1}$ on both sides, where for the first binomial, take the differentiated version of $(1+x)^r$ as its expansion, we then yield the sum:

$\sum_{k=0}^r k \binom{r}{k} \binom{m-r}{n-k} = \binom{m-1}{n-1}$

Hence the expected probability is:

$\frac{\binom{m-1}{n-1}}{\binom{m}{n}} = \frac{n}{m}$

harryharper · Sep 29, 2013

Re: HSC 2013 4U Marathon

Yep - just missing the 'r' from the RHS of your identity in the last two lines. Final answer should be nr/m.

harryharper · Sep 29, 2013

Re: HSC 2013 4U Marathon

$\text{Given that } \binom{n}{r}=0 $ for $ r>n $ or $ r<0,$

$Show that the sequence defined by$

$a_{n}=\sum_{k\ge0}\binom{k}{n-k}$

$is the sequence of Fibonnaci numbers.$

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