HSC 2015 MX1 Marathon (archive) (1 Viewer)

InteGrand · Jul 6, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Tjanks I didn't know that rule about cos (x +\alpha)) being equal to cos (x - \alpha)

This isn't true for all x, just some special x. Sy123 was saying $\cos (\pi +\alpha) = \cos (\pi -\alpha)$ .

We can also see this graphically as the cosine function's graph is symmetric about the line $x=\pi$ .

In fact, it is symmetric about any vertical line going through its peaks of troughs (i.e. +1 or -1). This is at $x = m\pi$ , for any integer $m$ , so we can say $\cos(m\pi +\alpha)=\cos(m\pi -\alpha)$ . A lot of similar identities can be derived for the cosine and related functions by looking at their graphs like this.

davidgoes4wce · Jul 6, 2015

Re: HSC 2015 3U Marathon

Need a bit of help to get this one started.

Sy123 · Jul 6, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Need a bit of help to get this one started.

Use the angle sum of a triangle being 180

We can see that in fact we know 2 angles, and these 2 angles are part of the same triangle, meaning .....

Speed6 · Jul 6, 2015

Re: HSC 2015 3U Marathon

X is also in a cylic quad meaning that once we do angle sum of a triangle for that top part, we can do 180- (180-70-30), do we agree?

We could also do co-interior, many ways in doing this question tbh, but it shouldn't take more than 2 lines.

Speed6 · Jul 6, 2015

Re: HSC 2015 3U Marathon

$$Show that$ ^{n+1}P_{k}=^{n}P_{k}+k$ ^{n}P_{k-1}$
$where$ ^{n}P_{k}=\frac{n!}{(n-k)!}$and n and k are positive integers with k$ < $n$

3 MARKS

DatAtarLyfe · Jul 6, 2015

Re: HSC 2015 3U Marathon

Speed6 said:
X is also in a cylic quad meaning that once we do angle sum of a triangle for that top part, we can do 180- (180-70-30), do we agree?

We could also do co-interior, many ways in doing this question tbh, but it shouldn't take more than 2 lines.

Use that. Then once you know that angle (lets call it alpha), alpha + X = 180 (opposite angles in a cyclic quadrilateral are supplementary)
Therefore X = 180 - alpha

rand_althor · Jul 6, 2015

Re: HSC 2015 3U Marathon

Speed6 said:
$$Show that$ ^{n+1}P_{k}=^{n}P_{k}+k$ ^{n}P_{k-1}$
$where$ ^{n}P_{k}=\frac{n!}{(n-k)!}$and n and k are positive integers with k$ < $n$

3 MARKS

$\begin{align*}\textup{RTP: }^{n+1}P_k&=^{n}P_k+k^{n}P_{k-1}\\\textup{RHS}&=\frac{n!}{(n-k)!}+\frac{n!k}{(n-k+1)!} \\&=\frac{n!(n-k+1)}{(n-k+1)!}+\frac{n!k}{(n-k+1)!} \\&=\frac{n!(n+1)-n!k+n!k}{(n-k+1)!} \\&=\frac{(n+1)!}{((n+1)-k)!} \\&=^{n+1}P_{k} \\&=\textup{LHS} \end{align*}$

leehuan · Jul 6, 2015

Re: HSC 2015 3U Marathon

Whilst we're here, bump.

leehuan said:
$Prove that $\left( \begin{matrix} n+1 \\ k+1 \end{matrix} \right) =\left( \begin{matrix} n \\ k \end{matrix} \right) +\left( \begin{matrix} n \\ k+1 \end{matrix} \right) \\ $1. By the definition of the binomial coefficient.$ \\ $2. Using a binomial expansion.$ \\ \\ $Explain why the result holds true with reference to Pascal's Triangle and combinations.$

Speed6 · Jul 6, 2015

Re: HSC 2015 3U Marathon

Nice setting out R_A

Drsoccerball · Jul 6, 2015

Re: HSC 2015 3U Marathon

leehuan said:
Whilst we're here, bump.

Using pascals triangle, We see how the two terms directly above a term add up to = to the term
$\therefore \binom{n+1}{k+1} =\binom{n}{k} + \binom{n}{k+1}$

RealiseNothing · Jul 6, 2015

Re: HSC 2015 3U Marathon

Speed6 said:
$$Show that$ ^{n+1}P_{k}=^{n}P_{k}+k$ ^{n}P_{k-1}$
$where$ ^{n}P_{k}=\frac{n!}{(n-k)!}$and n and k are positive integers with k$ < $n$

3 MARKS

Take an arbitrary element and don't choose it. The you get nPk of choosing rest.

Now choose that arbitrary element, you get latter term.

Speed6 · Jul 6, 2015

Re: HSC 2015 3U Marathon

RealiseNothing said:
Take an arbitrary element and don't choose it. The you get nPk of choosing rest.

Now choose that arbitrary element, you get latter term.

Show it, don't tell me

RealiseNothing · Jul 6, 2015

Re: HSC 2015 3U Marathon

Speed6 said:
Show it, don't tell me

My answer is fine.

Drsoccerball · Jul 6, 2015

Re: HSC 2015 3U Marathon

2. $(x+1)^{n+1}= (x+1)(nC0 +nc1 x +.... +(nCn)c^n)$
Compare coeficients of x^{k+1}
$\binom{n+1}{k+1} = \binom{n}{k} + \binom{n}{k+1}$

Drsoccerball · Jul 6, 2015

Re: HSC 2015 3U Marathon

RealiseNothing said:
My answer is fine.

I think he's looking for an algebraic approach

Drsoccerball · Jul 6, 2015

Re: HSC 2015 3U Marathon

$$ Show that : $ \\\ \binom{2n}{0}^2 -\binom{2n}{1}^2+\binom{2n}{2}^2 -...+\binom{2n}{2n}^2 = (-1)^n \cdto \binom{2n}{n}$

leehuan · Jul 6, 2015

Re: HSC 2015 3U Marathon

${ \left( 1-x \right) }^{ 2n }{ \left( 1+x \right) }^{ 2n }={ \left( 1-{ x }^{ 2 } \right) }^{ 2n }\\ Coefficient\quad on\quad { x }^{ 2n }={ \left( { x }^{ 2 } \right) }^{ n }:\\ RHS=\left( \begin{matrix} 2n \\ n \end{matrix} \right) { \left( -1 \right) }^{ n }\\ \left( \begin{matrix} c \\ r \end{matrix} \right) =\left( \begin{matrix} c \\ c-r \end{matrix} \right) ,\quad { \left( -1 \right) }^{ 2a }={ 1 }^{ a }=1$

$LHS=\left( \begin{matrix} 2n \\ 0 \end{matrix} \right) \left( \begin{matrix} 2n \\ 2n \end{matrix} \right) +\left( \begin{matrix} 2n \\ 1 \end{matrix} \right) \left( -1 \right) \left( \begin{matrix} 2n \\ 2n-1 \end{matrix} \right) +\left( \begin{matrix} 2n \\ 2 \end{matrix} \right) { \left( -1 \right) }^{ 2 }\left( \begin{matrix} 2n \\ 2n-2 \end{matrix} \right) +\dots +\left( \begin{matrix} 2n \\ 2n \end{matrix} \right) \left( \begin{matrix} 2n \\ 0 \end{matrix} \right) { \left( -1 \right) }^{ 2n }\\ ={ \left( \begin{matrix} 2n \\ 0 \end{matrix} \right) }^{ 2 }-{ \left( \begin{matrix} 2n \\ 1 \end{matrix} \right) }^{ 2 }+{ \left( \begin{matrix} 2n \\ 2 \end{matrix} \right) }^{ 2 }-\cdots +{ \left( \begin{matrix} 2n \\ 2n \end{matrix} \right) }^{ 2 }\\ Equate$

Drsoccerball · Jul 6, 2015

Re: HSC 2015 3U Marathon

you can also use:
$(1+x)^{2n}(1-\frac{1}{x})^{2n} =(x-\frac{1}{x})^{2n}$

Sy123 · Jul 6, 2015

Re: HSC 2015 3U Marathon

$\\ $i) Show that$ \ \frac{(1+x)^{n+1} -1}{n+1} = \sum_{k=0}^n \frac{1}{k+1} \binom{n}{k}x^{k+1}$

$\\ $ii) Hence show that$ \ \sum_{k=0}^n \frac{1}{(k+1)(n-k+1)} \binom{n}{k} \binom{n}{n-k} = \frac{1}{(n+1)^2} \binom{2n+2}{n+2}$

Crisium · Jul 6, 2015

Re: HSC 2015 3U Marathon

HSC 2015 MX1 Marathon (archive) (1 Viewer)

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