# Thread: Cambridge HSC MX1 Textbook Marathon/Q&A

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

Originally Posted by davidgoes4wce
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_...nts_pascal.htm
• Pascal's Triangle: http://www.mathwords.com/p/pascal's_triangle.htm .

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

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

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

• Binomial Coefficients in Pascal's Triangle: http://www.mathwords.com/b/binomial_...nts_pascal.htm
• Pascal's Triangle: http://www.mathwords.com/p/pascal's_triangle.htm .
$My thinking is ^nC_2+^{n+1}C_2=n^2 represents the value for the 3rd column$

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

3. ## 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.

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

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

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

5. ## 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.

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

Originally Posted by davidgoes4wce
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.

7. ## 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)

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

I'm not the most knowledgeable.

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

Originally Posted by InteGrand
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

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

Originally Posted by davidgoes4wce
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

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

Originally Posted by si2136
He probably did Uni maths and physics
Yeah that's what I thought lol, I wish there was a chemistry version of him

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

Originally Posted by eyeseeyou
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)

13. ## 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.

14. ## 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:

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

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

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

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

Originally Posted by davidgoes4wce
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 10th 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 "kth 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).

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

Originally Posted by Blitz_N7

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

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

Originally Posted by InteGrand
When we expand a 10th 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 "kth 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?

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

Originally Posted by davidgoes4wce
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.

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

Originally Posted by leehuan
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 .

20. ## 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

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

Originally Posted by leehuan
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.
Alot of the introductory problems with Binomials they want to get you to write the whole expansion out by using Binomial Theorem or Pascal's Triangle. Once you develop better understanding and knowledge, obviously you look at quicker methods.

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

Originally Posted by davidgoes4wce
In Question 13 (b) they ask for the 2 middle terms eg (a+3b)^5
If they ask for two then use what you said about sixth and seventh term for the higher power. But it's a bit more unlikely to be asked in the exam.

Originally Posted by davidgoes4wce
Alot of the introductory problems with Binomials they want to get you to write the whole expansion out by using Binomial Theorem or Pascal's Triangle. Once you develop better understanding and knowledge, obviously you look at quicker methods.
Do you really need to do more than 5 questions that make you write out a full binomial expansion to see what's going on?

Plus Pascal's triangle isn't really necessary. It helps to visualise things such as Pascal's identity but other than that so long as you know you can replace the terms with \binom{n}{k} nothing to worry about

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

On the topic of binomial stuff

$Prove that: \sum_{k=0}^n \binom{m+k}{m}=\binom{m+n+1}{m+1}.$

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

Originally Posted by seanieg89
On the topic of binomial stuff

$Prove that: \sum_{k=0}^n \binom{m+k}{m}=\binom{m+n+1}{m+1}.$
I think this is how to do that one

$\text{Compare the coefficient of }x^{m}\text{ in the expansion of }\\ (1+x)^m+(1+x)^{m+1}+\dots+(1+x)^{m+n}=\frac{(1+x)^ {m+n+1}}{x}-\frac{1}{x}$

$\text{Note that the coefficient of }x^m\text{ in the RHS is equivalent to the coefficient of }x^{m+1}\text{ in the numerator.}$

Edit: WOAH THERE. Start from (1+x)0, not (1+x)m otherwise the series is wrong

Result is not affected as there is no term in xm in the expansion of (1+x)k, where 0 ≤ k ≤ m-1

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

Yr 12 Cambridge Ex 10 A Q 23.

I spent more than 30 minutes and still couldn't come to the answer for this one. I had a few attempts at drawing the diagram but still couldn't come up with it.

Q23. A rectangular field is 60 metres long and 30 metres wide. A cow wanders randomly around the field. Find the probability that the cow is:

(a) More than 10 metres from the edge of the field

(b) not more than 10 metres from the corner of the field

Answers are:

$(a) \frac{2}{9}$

$(b) \frac{\pi}{18}$

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