Cambridge HSC MX1 Textbook Marathon/Q&A (1 Viewer)

InteGrand · Jun 8, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
Bump

"verify the result on the third column of the Pascal triangle." Does anyone have any idea on this? Or is my answer shown above good enough ?

$\noindent What it means is check up values in the third column of various rows of the Pascal triangle and see that the result holds.$

$\noindent $\Big{(}$Students are expected to know that the Pascal triangle entries are binomial coefficients. See the sites linked below. The first column in any row is better called column $0$ because it corresponds to $k=0$ in a binomial coefficient $\binom{n}{k}$. The top-most row is best called row 0 because it corresponds to $n=0$ for $\binom{n}{k}$. Using these conventions, the entry in row $n$ and column $k$ is $\binom{n}{k}$.$\Big{)}$$

• Binomial Coefficients in Pascal's Triangle: http://www.mathwords.com/b/binomial_coefficients_pascal.htm
• Pascal's Triangle: http://www.mathwords.com/p/pascal's_triangle.htm .

davidgoes4wce · Jun 8, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
$\noindent What it means is check up values in the third column of various rows of the Pascal triangle and see that the result holds.$

$\noindent $\Big{(}$Students are expected to know that the Pascal triangle entries are binomial coefficients. See the sites linked below. The first column in any row is better called column $0$ because it corresponds to $k=0$ in a binomial coefficient $\binom{n}{k}$. The top-most row is best called row 0 because it corresponds to $n=0$ for $\binom{n}{k}$. Using these conventions, the entry in row $n$ and column $k$ is $\binom{n}{k}$.$\Big{)}$$

• Binomial Coefficients in Pascal's Triangle: http://www.mathwords.com/b/binomial_coefficients_pascal.htm
• Pascal's Triangle: http://www.mathwords.com/p/pascal's_triangle.htm .

$My thinking is $ ^nC_2+^{n+1}C_2=n^2 $ represents the value for the 3rd column$

$Does 3rd column mean the same thing as k=2 ? if we take your assumption that { ^n\choose k}$

davidgoes4wce · Jun 8, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Im aware that Pascal's triangle is:

1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

I'm also aware there is no 3rd column in the first row.

InteGrand · Jun 8, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
$My thinking is $ ^nC_2+^{n+1}C_2=n^2 $ represents the value for the 3rd column$

$Does 3rd column mean the same thing as k=2 ? if we take your assumption that { ^n\choose k}$

Yes, k = 2.

davidgoes4wce · Jun 8, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

And what is the purpose of n?

If you look at Pascal's Triangle, the 3rd column values are: 1,3,6,10,15...........I struggle to see any link between that equation and those values.

InteGrand · Jun 8, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
And what is the purpose of n?

If you look at Pascal's Triangle, the 3rd column values are: 1,3,6,10,15...........I struggle to see any link between that equation and those values.

Pascal's Triangle:

1
1 1
1 2 1 (n = 2)
1 3 3 1 (n = 3)
1 4 6 4 1 (n = 4)
1 5 10 10 5 1 (n = 5)
etc.

The rule was that ⁿC₂ + _n+1C₂ = n².

In terms of the blue numbers above, the sum of consecutive ones will be n², where n is therow of the upper one out of the two.

E.g. 6 + 10 = 16 = 4² in the n = 4 row.

davidgoes4wce · Jun 8, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

OK I see it now.

Out of curiosity Integrand, I was keen to know what was your educational background? Your quite advanced in terms of knowledge and I think it's fair to say your the most knowledgeable on the forum. (I think others would agree)

InteGrand · Jun 8, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

I'm not the most knowledgeable.

eyeseeyou · Jun 8, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
I'm not the most knowledgeable.

he seems knowledgeable in maths and physics and not so much chemistry as he couldn't really answer my chem questions

si2136 · Jun 8, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
OK I see it now.

Out of curiosity Integrand, I was keen to know what was your educational background? Your quite advanced in terms of knowledge and I think it's fair to say your the most knowledgeable on the forum. (I think others would agree)

He probably did Uni maths and physics

eyeseeyou · Jun 8, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

si2136 said:
He probably did Uni maths and physics

Yeah that's what I thought lol, I wish there was a chemistry version of him

si2136 · Jun 8, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

eyeseeyou said:
Yeah that's what I thought lol, I wish there was a chemistry version of him

Chem isn't that useful outside of hs compared to physics or maths, as there are a lot of pathways that go with them (all types of Engineering, Maths, Physics, Computer Science)

Blitz_N7 · Jun 10, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

I'm having a hard time finding a systematic way of doing parts (c), (d) and (e). Is it necessary to check every number and eliminate the doubles or is there a easier way to do it? Thanks in advance.

davidgoes4wce · Jul 5, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

I don't want to look up the answer in the back of the book. Want to discuss it on an open forum.

$Cambridge Year 12 Exercise 5D Q 13(a) (i)$

$Find the middle term when the terms in the following expansions are arranged in increasing powers of y.$

$(i) (2x-3y)^{10}$

$$ What exactly do they mean by the middle term? I'm assuming it means the middle terms the 5th and 6th terms in this case OR the term $ x^5 y^5$

Here is my working out:

$(2x-3y)^{10}=$

$^{10}C_0 \times (2x)^{10} +^{10}C_1 \times (2x)^9 \times (-3y)^1 + ^{10}C_2 \times (2x)^8 \times (-3y)^2 +^{10}C_3 \times (2x)^7 \times (-3y)^3 +^{10}C_4 \times (2x)^6 \times (-3y)^4 +^{10}C_5 \times (2x)^5 \times (-3y)^5 +^{10}C_6 \times (2x)^4 \times (-3y)^6 + ^{10}C_7 \times (2x)^3 \times (-3y)^7 +^{10}C_8 \times (2x)^2 \times (-3y)^8 +^{10}C_9 \times (2x)^1 \times (-3y)^9 +^{10}C_{10} \times (-3y)^{10}$

$=1024x^{10}-15,360x^9y+103,680x^8y^2-414,720x^7y^3+1,088,640x^6y^4-1,959,552x^5y^5+2,449,440x^4y^6-2,099,520x^3y^7+1,180,980x^2y^8-393,660xy^9+59,049y^{10}$

InteGrand · Jul 5, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
I don't want to look up the answer in the back of the book. Want to discuss it on an open forum.

$Cambridge Year 12 Exercise 5D Q 13(a) (i)$

$Find the middle term when the terms in the following expansions are arranged in increasing powers of y.$

$(i) (2x-3y)^{10}$

$$ What exactly do they mean by the middle term? I'm assuming it means the middle terms the 5th and 6th terms in this case OR the term $ x^5 y^5$

Here is my working out:

$(2x-3y)^{10}=$

$^{10}C_0 \times (2x)^{10} +^{10}C_1 \times (2x)^9 \times (-3y)^1 + ^{10}C_2 \times (2x)^8 \times (-3y)^2 +^{10}C_3 \times (2x)^7 \times (-3y)^3 +^{10}C_4 \times (2x)^6 \times (-3y)^4 +^{10}C_5 \times (2x)^5 \times (-3y)^5 +^{10}C_6 \times (2x)^4 \times (-3y)^6 + ^{10}C_7 \times (2x)^3 \times (-3y)^7 +^{10}C_8 \times (2x)^2 \times (-3y)^8 +^{10}C_9 \times (2x)^1 \times (-3y)^9 +^{10}C_{10} \times (-3y)^{10}$

$=1024x^{10}-15,360x^9y+103,680x^8y^2-414,720x^7y^3+1,088,640x^6y^4-1,959,552x^5y^5+2,449,440x^4y^6-2,099,520x^3y^7+1,180,980x^2y^8-393,660xy^9+59,049y^{10}$

When we expand a 10^th power using binomial expansion, there'll be 11 terms. So the middle term is the sixth term (as there will be five terms either side of this, for a total of 11 terms).

We don't need to expand everything to find just the sixth (or any particular) term. We can write down the general "k^th term" using the binomial theorem, and then substituting k = 5 gives us the term we want (reason to sub. in 5 is that k goes from 0 to 10, not from 1 to 11).

leehuan · Jul 5, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Blitz_N7 said:
I'm having a hard time finding a systematic way of doing parts (c), (d) and (e). Is it necessary to check every number and eliminate the doubles or is there a easier way to do it? Thanks in advance.

Divisible by 5 - consider the last digit first. It has to be 5 (because 0 is not an option). Then arrange everything else

Divisible by 3 - This is trickier. The SUM of the two numbers has to be divisible by 3.

Or you can play it smart. Because the numbers are forcibly 2 digits, then:
From 0-100 there are 33 numbers divisible by 3
From 0-9 there are 3 numbers divisible by 3
Then do subtraction

Divisible by 6 - I would've done this using set theory. Divisible by 6 just means both even and divisible by 3 simultaneously

davidgoes4wce · Jul 5, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
When we expand a 10^th power using binomial expansion, there'll be 11 terms. So the middle term is the sixth term (as there will be five terms either side of this, for a total of 11 terms).

We don't need to expand everything to find just the sixth (or any particular) term. We can write down the general "k^th term" using the binomial theorem, and then substituting k = 5 gives us the term we want (reason to sub. in 5 is that k goes from 0 to 10, not from 1 to 11).

Yep I know there are 11 terms if it is of the form $(ax+b)^{10}$

$$ I also know the formula for the term is $ T_{k+1}=^nC_k \ (a)^{n-k} \ b^k$

$$ Applying to this for the case $ (2x-3y)^{10}$

$So T_{6}=^{10}C_5 \ (2x)^{10-5} \ (-3y)^{5} =$

I personally think its good to do it both ways 1) the long way and 2) the shorter way for learning purposes.

Food for thought, what happens if the term was to the power of 11 ?

i.e $(ax+b)^{11}$ ? Would we then look at the 6th term AND 7th term as the middle term?

leehuan · Jul 5, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
Yep I know there are 11 terms if it is of the form $(ax+b)^{10}$

$$ I also know the formula for the term is $ T_{k+1}=^nC_k \ (a)^{n-k} \ b^k$

$$ Applying to this for the case $ (2x-3y)^{10}$

$So T_{6}=^{10}C_5 \ (2x)^{10-5} \ (-3y)^{5} =$

I personally think its good to do it both ways 1) the long way and 2) the shorter way for learning purposes.

Food for thought, what happens if the term was to the power of 11 ?

i.e $(ax+b)^{11}$ ? Would we then look at the 6th term AND 7th term as the middle term?

Unnecessarily expanding feels like a waste of time and does not teach much at all.

They wouldn't ask you for a middle term that way, they will either make it plural (to which therefore yes) or just not ask it.

davidgoes4wce · Jul 5, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

leehuan said:
Unnecessarily expanding feels like a waste of time and does not teach much at all.

They wouldn't ask you for a middle term that way, they will either make it plural (to which therefore yes) or just not ask it.

OK Cool, what I did notice with those questions were they were asking them to even powers .

davidgoes4wce · Jul 5, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

In Question 13 (b) they ask for the 2 middle terms eg (a+3b)^5

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