HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Sep 29, 2013

Re: HSC 2013 4U Marathon

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

$Show that the sequence defined by$

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

$is the sequence of Fibonnaci numbers.$

$a_{2m-1} + a_{2m-2} = \binom{m-1}{m-1} + \left(\binom{m}{m-1} + \binom{m}{m-2} \right) + \left( \binom{m+1}{m-2} + \binom{m+1}{m-3} \right) + \dots + \left(\binom{2m-2}{0} + \binom{2m-2}{1} \right ) + \binom{2m-1}{0}$

$Use the identity$

$\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$

$$and$ \ \ \binom{2m-1}{0} = \binom{2m}{0}$

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

$\therefore \ \ a_{2m-2} + a_{2m-1} = a_{2m}$

Do the same for n = 2k+1, and we then prove that:

$a_n = a_{n-1} + a_{n-2}$ for all n, therefore it is the sequence of Fibonacci numbers.

-----

$$Assume$ \ \ \lim_{x \to 0} \sqrt[x]{1+x} = e$

$Define$

$s_n = \prod_{k=0}^n \binom{n}{k}$

$\lim_{n \to \infty} \frac{\left(\frac{s_{n+1}}{s_n} \right)}{\left(\frac{s_{n}}{s_{n-1}} \right)} = e$

harryharper · Sep 29, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Assume$ \ \ \lim_{x \to 0} \sqrt[x]{1+x} = e$

$Define$

$s_n = \prod_{k=0}^n \binom{n}{k}$

$\lim_{n \to \infty} \frac{\left(\frac{s_{n+1}}{s_n} \right)}{\left(\frac{s_{n}}{s_{n-1}} \right)} = e$

$\frac{s_n}{s_{n-1}}=\binom{n}{n}\prod_{k=0}^{n-1}\frac{\binom{n}{k}}{\binom{n-1}{k}}=\prod_{k=0}^{n-1}\frac{n}{n-k}=\frac{n^n}{n!}$

$\therefore \frac{\left(\frac{s_{n+1}}{s_n} \right)}{\left(\frac{s_{n}}{s_{n-1}} \right)}=\frac{\left(\frac{(n+1)^{n+1}}{(n+1)!} \right)}{\left(\frac{n^n}{n!} \right)}=\left(\frac{n+1}{n}\right)^n$

$\text{and from the given limit, it is clear that }\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e\text{ as required.}$

psychotropic · Sep 30, 2013

Re: HSC 2013 4U Marathon

Three distinct numbers a,b,c are chosen at random from the numbers 1,2,...100. Find the probability that a,b and c are in a geometric progression

Sy123 · Sep 30, 2013

Re: HSC 2013 4U Marathon

psychotropic said:
Three distinct numbers a,b,c are chosen at random from the numbers 1,2,...100. Find the probability that a,b and c are in a geometric progression

If its in geometric progression, then for integers a and k, it must be in the form:

$a, ak, ak^2$

Essentially, we are finding how many integer solutions there are for:

$ak^2 \leq 100$

Where (a,k) are positive integers with $a \leq 100$

So lets sketch the graph:

$y^2 = \frac{100}{x}$ for positive y, therefore:

$y = \frac{10}{\sqrt{x}}$

When we sketch this graph, we have to find all integer points lying on or under the curve.
To measure this we sketch the lines, y=1, y=2, y=3, y=4 ... y=10

The intersection of y=1 and the curve is (100,1), that means (99,1) and (98,1) and so on are under the curve.
That is 100 solutions EDIT: oops

y=2, (25,2), that is another 25 solutions

y=3, (100/9, 3), that is 11 solutions

y=4, (6.25, 4) that is 6 soln

y=5, (4, 5), that is 4 soln

y=6, (2.7, 6), that is 2 soln

y=7, (2.0, 7) that is 2

y=8 (1. , 8) that is 1

y=9 (1.,, , 9) that is 1

y=10 (1, 10) that is 1.

Add them all together to get 153 integer pair solutions, that means the probability of picking a geometric progression is:

$\frac{153}{C(100,3)} = \frac{51}{53900}$ ???

Sy123 · Sep 30, 2013

Re: HSC 2013 4U Marathon

$$You are given that for a certain radius of convergence, a function$ \ \ f(x) \ \ $which is infinitely differentiable at$ \ \ x=0$

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)} (0)}{n!}x^n$

$i) Show that$

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$

$ii) Given that$

$\sin^{-1} x = \sum_{n=0}^{\infty} c_n x^n$

$Show that$ \ \ c_n \ \ $satisfies$

$x= \sum_{n=0}^{\infty} c_n \left(\sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!} \right )^n$

$iii) By differentiating part (ii) (which you may assume is possible) Show that$

$\sin^{-1} x = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{4^n (2n+1)}x^{2n+1}$

psychotropic · Sep 30, 2013

Re: HSC 2013 4U Marathon

pretty much except u have to discard those 100 solutions since we have to have distinct a,b,c. Otherwise you are correct.

seanieg89 · Sep 30, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$You are given that for a certain radius of convergence, a function$ \ \ f(x) \ \ $which is infinitely differentiable at$ \ \ x=0$

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)} (0)}{n!}x^n$

$i) Show that$

$\sin x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots$

$ii) Given that$

$\sin^{-1} x = \sum_{n=0}^{\infty} c_n x^n$

$Show that$ \ \ c_n \ \ $satisfies$

$x= \sum_{n=0}^{\infty} c_n \left(\sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!} \right )^n$

$iii) By differentiating part (ii) (which you may assume is possible) Show that$

$\sin^{-1} x = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{4^n (2n+1)}x^{2n+1}$

A bit flawed, the radius of convergence of a Taylor series can be 0. Eg consider the function defined piecewise as e^{-1/x} for x > 0 and 0 for x < 0.

sine just happens to have an infinite radius of convergence.

harryharper · Sep 30, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
If its in geometric progression, then for integers a and k, it must be in the form:

$a, ak, ak^2$

Essentially, we are finding how many integer solutions there are for:

$ak^2 \leq 100$

Where (a,k) are positive integers with $a \leq 100$

Isn't this not quite right?

It's not specified that the common ratio must be an integer. Here we're missing out on solutions like (4,6,9), (25,30,36), (27,36,48), (44,66,99)...In fact, there are nearly twice as many solutions (I think) if we allow non-integer ratios.

That is, if we solve instead for $ap^2, apq, aq^2$ , where $GCD(p,q)=1$ and $p<q$

A very similar approach gives 53 solutions for p=1 (the case that Sy solved), and 52 more for other values of p. Instead of solving for #values of k under $\frac{10}{\sqrt{a}}$ , we're solving for the number of coprime pairs.

31 for a=1, 17 for a=2, 9 for each of a=3 and a=4, 5 for each of a=5,6, 3 for each of a=7,8,9,10,11 and 1 for each of a=12,13,14,...,25

31 + 17 + 2x9 + 2x5 + 5x3 + 14x1 = 105

Sy123 · Sep 30, 2013

Re: HSC 2013 4U Marathon

harryharper said:
Isn't this not quite right?

It's not specified that the common ratio must be an integer. Here we're missing out on solutions like (4,6,9), (25,30,36), (27,36,48), (44,66,99)...In fact, there are nearly twice as many solutions (I think) if we allow non-integer ratios.

That is, if we solve instead for $ap^2, apq, aq^2$ , where $GCD(p,q)=1$ and $p<q$

A very similar approach gives 53 solutions for p=1 (the case that Sy solved), and 52 more for other values of p. Instead of solving for #values of k under $\frac{10}{\sqrt{a}}$ , we're solving for the number of coprime pairs.

31 for a=1, 17 for a=2, 9 for each of a=3 and a=4, 5 for each of a=5,6, 3 for each of a=7,8,9,10,11 and 1 for each of a=12,13,14,...,25

31 + 17 + 2x9 + 2x5 + 5x3 + 14x1 = 105

Very true, my bad.

Sy123 · Oct 2, 2013

Re: HSC 2013 4U Marathon

$Find a recurrence formula for$

$I(m,k,n) = \int_0^1 x^m (1-x^k)^n \ dx$

$Hence prove$

$\left( \frac{k}{m+n+1} \right)^n < \frac{1}{n!} \int_0^1 x^m (1-x^k)^n \ dx < \left( \frac{k}{m+1} \right)^n$

Sy123 · Oct 2, 2013

Re: HSC 2013 4U Marathon

$A projectile is fired from the origin with a velocity of$ \ V \ \ $to hit a wall$ \ \ d \ \ $units away from the origin$

$$Ignore air resistance, take gravitational acceleration as$ \ \ g$

$Show that if it can reach the wall above ground level if and only if$

$V > \sqrt{gd}$

JJ345 · Oct 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$A projectile is fired from the origin with a velocity of$ \ V \ \ $to hit a wall$ \ \ d \ \ $units away from the origin$

$$Ignore air resistance, take gravitational acceleration as$ \ \ g$

$Show that if it can reach the wall above ground level if and only if$

$V > \sqrt{gc}$

The letters in the square root are supposed to be gd right?
t = d/(Vcosa) --> where a is the varying angle
Sub into y = -1/2 gt^2 + Vtsina
And let a=45 (minimum conditions for the projectile to hit the wall as range is max at this angle so velocity will the be min required--> I've worded this terribly

)
y=d-gd^2/v^2
y>0 for it to hit the wall
So rearranging, v>sqrt(gd)

Sy123 · Oct 2, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
The letters in the square root are supposed to be gd right?
t = d/(Vcosa) --> where a is the varying angle
Sub into y = -1/2 gt^2 + Vtsina
And let a=45 (minimum conditions for the projectile to hit the wall as range is max at this angle so velocity will the be min required--> I've worded this terribly )
y=d-gd^2/v^2
y>0 for it to hit the wall
So rearranging, v>sqrt(gd)

Yeah I think that makes sense, well done.

My way was more straightforward, by subbing in t=d/Vcos a, and then y > 0
We arrive at:

$V^2 > \frac{gd}{2} (\tan \theta + \cot \theta)$

Since (a + 1/a) > 2, put in a = than theta to get, $\tan \theta + \cot \theta \geq 2$

JJ345 · Oct 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yeah I think that makes sense, well done.

My way was more straightforward, by subbing in t=d/Vcos a, and then y > 0
We arrive at:

$V^2 > \frac{gd}{2} (\tan \theta + \cot \theta)$

Since (a + 1/a) > 2, put in a = than theta to get, $\tan \theta + \cot \theta \geq 2$

From your second last step if we rearrange in terms of tan theta then solve for discriminant we get the required expression aswell--> But what that means physically doesn't really make sense to me--> Is it just that yes there's a real value for tan theta so its hitting the wall at some angle??

Sy123 · Oct 2, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
From your second last step if we rearrange in terms of tan theta then solve for discriminant we get the required expression aswell--> But what that means physically doesn't really make sense to me--> Is it just that yes there's a real value for tan theta so its hitting the wall at some angle??

The inequality came about due to the condition that y > 0 at x = d, thus if we take a quadratic in terms of tan theta, it will describe all possible tan theta in terms of V such that y > 0 at x=d (i.e. the conditions we need for the question), so yeah that's another method

harryharper · Oct 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Practising for the BOS trial? (btw that was HSC 2007 4U 13 marks compressed).

$You may use the fact that$

$\sum_{k=1}^{\infty} \frac{\sin(k\theta)}{k} = \frac{\pi}{2} - \frac{\theta}{2}$

$Prove that$

$\pi = 3 + \frac{4}{2 \cdot 3 \cdot 4} - \frac{4}{4\cdot 5 \cdot 6} + \frac{4}{6 \cdot 7 \cdot 8} - \frac{4}{8\cdot 9 \cdot 10} + \dots$

$\theta=\pi/2\text{ yields }\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots$

$\text{We know from partial fractions that } \frac{1}{(4n-2)(4n-1)4n}-\frac{1}{4n(4n+1)(4n+2)} = \frac{1}{4}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+\frac{1}{4n+1}-\frac{1}{4n-1}$

$\text{Also, }\sum_{k=1}^{\infty} \frac{1}{4}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+\frac{1}{4n+1}-\frac{1}{4n-1}= \frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{2n-1}-\frac{1}{2n+1}+\sum_{k=1}^{\infty} \frac{1}{4n+1}-\frac{1}{4n-1}$

$\text{But }\sum_{k=1}^{\infty}\frac{1}{2n-1}-\frac{1}{2n+1}=1\text{ by telescoping and }\sum_{k=1}^{\infty} \frac{1}{4n+1}-\frac{1}{4n-1}=-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots=\frac{\pi}{4}-1$

$\therefore 3 + \frac{4}{2 \cdot 3 \cdot 4} - \frac{4}{4\cdot 5 \cdot 6} + \frac{4}{6 \cdot 7 \cdot 8} - \frac{4}{8\cdot 9 \cdot 10} + \dots = 3 + 4\left(\frac{1}{4}+\frac{\pi}{4}-1\right)=\pi\text{ as required.}$

Sy123 · Oct 2, 2013

Re: HSC 2013 4U Marathon

harryharper said:
$\theta=\pi/2\text{ yields }\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots$

$\text{We know from partial fractions that } \frac{1}{(4n-2)(4n-1)4n}-\frac{1}{4n(4n+1)(4n+2)} = \frac{1}{4}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+\frac{1}{4n+1}-\frac{1}{4n-1}$

$\text{Also, }\sum_{k=1}^{\infty} \frac{1}{4}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+\frac{1}{4n+1}-\frac{1}{4n-1}= \frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{2n-1}-\frac{1}{2n+1}+\sum_{k=1}^{\infty} \frac{1}{4n+1}-\frac{1}{4n-1}$

$\text{But }\sum_{k=1}^{\infty}\frac{1}{2n-1}-\frac{1}{2n+1}=1\text{ by telescoping and }\sum_{k=1}^{\infty} \frac{1}{4n+1}-\frac{1}{4n-1}=-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots=\frac{\pi}{4}-1$

$\therefore 3 + \frac{4}{2 \cdot 3 \cdot 4} - \frac{4}{4\cdot 5 \cdot 6} + \frac{4}{6 \cdot 7 \cdot 8} - \frac{4}{8\cdot 9 \cdot 10} + \dots = 3 + 4\left(\frac{1}{4}+\frac{\pi}{4}-1\right)=\pi\text{ as required.}$

Yep well done.

Can anyone post a question please? (preferably probability)

VBN2470 · Oct 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yep well done.

Can anyone post a question please? (preferably probability)

An unbiased die is thrown six times. Find the probability that the six scores obtained will be such that a 6 occurs only on the last throw and exactly three of the first five throws result in odd numbers.

VBN2470 · Oct 3, 2013

Re: HSC 2013 4U Marathon

Prove that $\frac{(2m)!(2n)!}{(m+n)!m!n!}$ is an integer for all positive integer values of $\text m, n$ .

Sy123 · Oct 3, 2013

Re: HSC 2013 4U Marathon

VBN2470 said:
Prove that $\frac{(2m)!(2n)!}{(m+n)!m!n!}$ is an integer for all positive integer values of $\text m, n$ .

Proving by mathematical induction on n, for integer m

$m \neq n$

$$Step 1$ \ \ n=1$

$\frac{(2m)! \times 2}{(m+1)! m!} = \frac{2}{m+1} \binom{2m}{m}$

$= 2\left(\binom{2m}{m} - \binom{2m}{m+1} \right) = $integer$$

$$Step 2$ \ \ n=k$

$$Assume for integer$ \ \ p$

$p = \frac{(2k)! (2m)!}{(m+k)! m! k!}$

$$Step 3$ \ \ n=k+1 \ \ $we must prove for integer$ \ \ q$

$q = \frac{(2k+2)! (2m)!}{(m+k+1)! m! (k+1)!}$

$q = p \frac{2(2n+1)}{(m+n+1)}$

We know that

$m < n$ OR

$m > n$

Therefore, $m+n+1 \leq 2m$ OR $m+n+1 \leq 2n$

$$Therefore$ \ \ (2n)! \ \ $or$ \ \ (2m)! \ \ $has a factor of$ \ \ (m+n+1)$

$$i.e.$ \ \ (2n)! = (2n)(2n-1)\dots (n+m+1) \dots 1$

$\therefore \ \ p \ \ $is divisible by$ \ \ m+n+1$

$\therefore \ \frac{2(2k+1)}{m+k+1} p = q \ \ $is integer$$

For the case $m = n$

$\frac{(2n)!^2}{(2n)! n!^2} = \binom{2n}{n} = $integer$$

$Therefore the expression is always an integer$

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