• 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 .
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.
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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.
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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 ^{n}C_{2} + _{n+1}C_{2} = n^{2}.
In terms of the blue numbers above, the sum of consecutive ones will be n^{2}, where n is therow of the upper one out of the two.
E.g. 6 + 10 = 16 = 4^{2} in the n = 4 row.
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)
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I'm not the most knowledgeable.
USYD 2018: MAJ: FINC/LAW MIN: MATH/PHIL
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.
I don't want to look up the answer in the back of the book. Want to discuss it on an open forum.
Here is my working out:
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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).
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
Yep I know there are 11 terms if it is of the form
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 ? Would we then look at the 6th term AND 7th term as the middle term?
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In Question 13 (b) they ask for the 2 middle terms eg (a+3b)^5
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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.
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
On the topic of binomial stuff
Last edited by leehuan; 5 Jul 2016 at 4:59 PM.
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:
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