WEBVTT

4ffabaca-4711-48a5-aff6-8f1719b31bcf-0
00:00:05.400 --> 00:00:06.600
Hi, my name's Richard.

0dde3b1f-f6a1-4f82-b423-8aee107f9d09-0
00:00:06.640 --> 00:00:10.680
We're going to be looking at Tamua 2021
Paper one, question 16.

28f22a2f-4e71-43a6-ad1f-a349fc0b6984-0
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Here's the question.

89b7df36-c751-4c44-aac4-09bd3fc73aab-0
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It says consider the expansion of A plus
BX to the north.

27e9f055-e013-4721-8c65-799453bd5e91-0
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We've got some information about it.

b4b3316a-af37-4730-a5ab-d3d38c9e3589-0
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The third term in ascending powers of X
is 105 X squared.

b9f4aa07-ca59-48e0-8022-4a32958327da-0
00:00:23.160 --> 00:00:27.792
The fourth term in ascending powers of X
is 210 X cubed,

b9f4aa07-ca59-48e0-8022-4a32958327da-1
00:00:27.792 --> 00:00:33.400
and the fourth term notice in descending
powers of X is 210 X cubed.

83c7d3f3-62c7-472b-aaa2-4d42e65f4e32-0
00:00:33.760 --> 00:00:37.280
And we are asked to find the value of a /
b ^2.

49aa14ad-4f6b-450b-901d-a5b7e247f1ba-0
00:00:37.880 --> 00:00:43.161
First thing to note about this is that
this means what we get when we take a / b

49aa14ad-4f6b-450b-901d-a5b7e247f1ba-1
00:00:43.161 --> 00:00:44.400
and then square it.

2b39cc18-77ee-4b72-8247-3b450e1198e1-0
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So perhaps it will be helpful for us to
think of a bracket here just to emphasise

2b39cc18-77ee-4b72-8247-3b450e1198e1-1
00:00:48.959 --> 00:00:52.240
that that's what we're looking for,
the value of a / b ^2.

6243c715-35a2-4514-834e-185029aca8b8-0
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OK.

4c6e27cf-5106-4d2a-9593-5fb66025f5b5-0
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The first thing we're going to do is to
think about how this information tells us

4c6e27cf-5106-4d2a-9593-5fb66025f5b5-1
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how many terms there are in our binomial
expansion.

93224103-ac85-4776-b254-8fe4ee85c03e-0
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So that will tell us the exact value of N
if we imagine expanding A+, BX to the N.

a5d170c1-308f-4a3e-a1ee-23ed5efd827b-0
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So when we do this,
we create a number of terms.

c478c9a4-0c23-4c86-82d5-0d3005893547-0
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We'll have our first term,
which would just be A to the N,

c478c9a4-0c23-4c86-82d5-0d3005893547-1
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just a number with no XS.

902394d3-3917-4832-a994-81f89b779b3c-0
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We would have another term with an X,
the X term,

902394d3-3917-4832-a994-81f89b779b3c-1
00:01:21.432 --> 00:01:27.198
another term the third term with the X ^2,
and we're told that this one is this one

902394d3-3917-4832-a994-81f89b779b3c-2
00:01:27.198 --> 00:01:28.160
105 X squared.

e6c980ba-75a3-4486-bd0d-c6347dd01679-0
00:01:29.640 --> 00:01:32.880
So this is our third term in ascending
powers of X.

2704aa3a-40c3-4a4d-b0aa-2fdb02d77fe9-0
00:01:33.400 --> 00:01:39.480
The fourth term in ascending powers of X
is this 1, so 210 X cubed.

00275550-f9db-424b-9a54-99419cbd5de2-0
00:01:41.000 --> 00:01:46.120
And then we're told that the fourth term
in descending powers of X is 210 X cubed.

c32f091c-e4e2-4b65-adc9-c20684137eae-0
00:01:46.400 --> 00:01:48.742
Well,
there's only One X ^3 term in this

c32f091c-e4e2-4b65-adc9-c20684137eae-1
00:01:48.742 --> 00:01:51.485
expansion,
so that is another reference to this

c32f091c-e4e2-4b65-adc9-c20684137eae-2
00:01:51.485 --> 00:01:52.399
particular term.

2e4e1409-20cc-493f-8587-1a8ef83b9b63-0
00:01:52.840 --> 00:01:56.155
And we're being told that this is the
fourth one from the end,

2e4e1409-20cc-493f-8587-1a8ef83b9b63-1
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counting from the right back in.

e8d5ac36-9ee1-45fb-a768-639cf7622a9d-0
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So there must be another term after this
which would be the X to the fourth term,

e8d5ac36-9ee1-45fb-a768-639cf7622a9d-1
00:02:03.463 --> 00:02:08.442
and then another term after this which
would be the X to the five term and then

e8d5ac36-9ee1-45fb-a768-639cf7622a9d-2
00:02:08.442 --> 00:02:13.173
another term after this in order that
this one is the 4th one in descending

e8d5ac36-9ee1-45fb-a768-639cf7622a9d-3
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powers of X.

61a26e63-4e28-495f-865d-c25fd9ae25da-0
00:02:14.480 --> 00:02:18.200
So what that means is we would have an X
to the sixth term here.

9d478f64-5185-4eb0-a2bd-7448e5cadc96-0
00:02:18.880 --> 00:02:23.018
And that tells us immediately that the
value of N in this question is 6,

9d478f64-5185-4eb0-a2bd-7448e5cadc96-1
00:02:23.018 --> 00:02:27.440
in order that the highest power of X at
this end is is actually X to the six.

7c145d73-0dd2-4251-8b6f-82a295a71bf2-0
00:02:28.040 --> 00:02:31.200
So from this we can deduce that n = 6.

036f7719-e989-4672-9c47-6b85e1ea6a52-0
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And it's really coming from these two
pieces of information.

32c6c2eb-b510-4694-b199-4075faeffd81-0
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OK, so now we know that n = 6,
we can replace this N here with six,

32c6c2eb-b510-4694-b199-4075faeffd81-1
00:02:40.829 --> 00:02:46.351
and we can think about what the expansion
would look like for this particular value

32c6c2eb-b510-4694-b199-4075faeffd81-2
00:02:46.351 --> 00:02:46.680
of 6.

1150eca5-0544-44c4-a147-950e3ff6d5c3-0
00:02:47.120 --> 00:02:51.520
OK, so moving through the terms,
our first term would be 8 to the 6th.

e7d989e9-69f6-45cb-9cf4-779a94c2834f-0
00:02:52.400 --> 00:02:54.800
Our second term would be 6.

e3df484a-a5ee-4bc2-a989-8ff064f182f4-0
00:02:54.800 --> 00:03:00.875
Choose one, and now we have A to the five,
and we would have BX to the power of 1,

e3df484a-a5ee-4bc2-a989-8ff064f182f4-1
00:03:00.875 --> 00:03:02.120
so just BX there.

df8c6f7c-f7c9-4dce-b433-e1161e12298a-0
00:03:03.320 --> 00:03:05.640
Our third term would be 6.

05cfb192-573c-4a4a-9461-6a9f85293003-0
00:03:05.640 --> 00:03:08.993
Choose two,
which I'll write in this notation again,

05cfb192-573c-4a4a-9461-6a9f85293003-1
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we would now have A to the four and we
would have the square of this BX,

05cfb192-573c-4a4a-9461-6a9f85293003-2
00:03:13.612 --> 00:03:16.080
which I'm going to write as b ^2, X ^2.

d9528f0a-2ec6-4812-ac64-fa33f4efda40-0
00:03:16.520 --> 00:03:18.200
So this is actually our second term.

59e5d08e-64e1-41bd-847a-cd06133556fa-0
00:03:18.200 --> 00:03:20.880
We know this one is this one of 105 X.

e640703d-e2de-418b-ac84-e2ebf67b5476-0
00:03:21.320 --> 00:03:25.932
And our third term, sorry,
our fourth term in ascending powers of X

e640703d-e2de-418b-ac84-e2ebf67b5476-1
00:03:25.932 --> 00:03:27.560
would be 6 choose three.

b4977cea-2135-45dc-8cfe-c114c0c1f743-0
00:03:28.280 --> 00:03:35.720
Now we have a cubed and we also have the
cube of BX, which is b ^3 X ^3.

c4e0b268-a6d1-4c71-8dae-dad0cdda838f-0
00:03:36.160 --> 00:03:40.677
Now this is as far as I need to go to use
this information to deduce what we need

c4e0b268-a6d1-4c71-8dae-dad0cdda838f-1
00:03:40.677 --> 00:03:42.000
to deduce about A&B.

982a000c-01a3-4d64-8562-de22fcc59d5f-0
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OK,
so we have that six choose 2A to the four

982a000c-01a3-4d64-8562-de22fcc59d5f-1
00:03:46.805 --> 00:03:50.560
b ^2,
X ^2 is this term here 105 X squared.

f8a51508-8447-41a6-9b21-413f1e19bd3b-0
00:03:50.960 --> 00:03:54.040
So that tells us that six choose 2.

2889215a-dd4f-4066-9bc2-ce65161e8487-0
00:03:55.240 --> 00:03:59.920
A to the 4B squared is 105.

e594b00d-610b-467d-9e51-bbf087f6d602-0
00:04:02.480 --> 00:04:06.960
6 choose 2 means 6 * 5 / 2 * 1.

36d6a821-830d-46f7-b2bc-79d1ba219ccd-0
00:04:07.440 --> 00:04:09.560
So that's actually 30 / 2.

14f2f2a5-0dd5-4963-a77f-f01129738934-0
00:04:09.800 --> 00:04:11.200
This number is 15.

789872cd-68f7-4c14-8fa3-92cd5bf22332-0
00:04:11.720 --> 00:04:16.040
If I divide my 105 here by 15,
I will get 7.

cf741182-5c6c-4f3f-867b-d5451fec591f-0
00:04:16.480 --> 00:04:22.360
So this tells me that A to the four b ^2
= 7.

d8a1eadf-363f-4dc3-8a96-ce84ca6eed23-0
00:04:22.680 --> 00:04:27.280
We also have that this term here is 210 X
cubed.

8cef3c3c-7609-460f-869d-6657a39b0dff-0
00:04:27.520 --> 00:04:33.162
So we have that six choose 3 * a ^3 * b
^3 = 210,

8cef3c3c-7609-460f-869d-6657a39b0dff-1
00:04:33.162 --> 00:04:38.240
equating the coefficients of the X ^3
there.

ad030fdc-00c4-44c4-8963-e1aa294090f1-0
00:04:38.760 --> 00:04:47.800
Now 6 choose three means 6 * 5 * 4 / 3 *
2 * 1, that 3 * 2 * 1, that's a six.

69e20d71-8ab1-4094-afdc-d3fbe157c343-0
00:04:47.920 --> 00:04:51.080
So that cancels the six on the top and it
will just be 5 * 4.

615bdbd8-c828-49f5-b162-678333cb864c-0
00:04:51.480 --> 00:04:59.640
So this will become twenty a ^3 b ^3 =
210.

09f0a579-d9a9-4681-a10a-99a040e8b7c6-0
00:05:00.880 --> 00:05:08.600
And if I divide both sides by 20 here,
that will give me the a ^3, b ^3 equals.

5e5f7102-64e4-4aa6-b729-f3f64d789117-0
00:05:08.600 --> 00:05:12.040
Well, first of all I'll have 210 / 20.

4312b143-513a-437b-8969-75f0a8daddc4-0
00:05:12.320 --> 00:05:20.167
That's the same as 21 / 2 though I now
have the value of A to the 4B squared as

4312b143-513a-437b-8969-75f0a8daddc4-1
00:05:20.167 --> 00:05:24.680
7,
and also the value of a ^3 b ^3 as 21 / 2.

2bd7fbc3-bd73-4784-8c8f-54125e6c97c4-0
00:05:25.600 --> 00:05:29.440
I need to know the value of A / b all
squared.

ba4cc7ac-e613-460b-b4ba-9b3861a3426c-0
00:05:30.280 --> 00:05:36.584
And now I need to notice that if I were
to divide A to the 4B squared by a ^3 b

ba4cc7ac-e613-460b-b4ba-9b3861a3426c-1
00:05:36.584 --> 00:05:40.210
^3,
that would actually give me A / b because

ba4cc7ac-e613-460b-b4ba-9b3861a3426c-2
00:05:40.210 --> 00:05:43.520
I've got one more a here than I have here.

4a0d51a8-80a3-4dd7-83c2-6a9a67345d52-0
00:05:43.520 --> 00:05:47.056
So I'd leave an A on the top,
and I've got 1 fewer B here than I have

4a0d51a8-80a3-4dd7-83c2-6a9a67345d52-1
00:05:47.056 --> 00:05:49.280
here,
so that would leave AB on the bottom.

019797e0-0f57-4c74-9396-920f5d634821-0
00:05:49.640 --> 00:05:52.040
OK, so I'm going to do that now.

cb70d96c-46eb-48c9-9a98-d4cc7ac4a37f-0
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I'm going to divide A to the 4B squared
by a ^3 b ^3 so I can see from the

cb70d96c-46eb-48c9-9a98-d4cc7ac4a37f-1
00:05:58.138 --> 00:06:04.600
algebra that this will give me A over BI
have one more a on the top then on the

cb70d96c-46eb-48c9-9a98-d4cc7ac4a37f-2
00:06:04.600 --> 00:06:08.720
bottom,
one fewer B on the top than on the bottom.

f28cbd69-fda3-4260-bab3-7a7ab34d8607-0
00:06:08.720 --> 00:06:10.080
So this will give me A / b.

e1ffdf2c-be2f-4d72-ba44-9302bbe303e6-0
00:06:10.440 --> 00:06:12.080
But I also know these values.

bee6681f-f738-4f4d-b1ff-26df46e6a506-0
00:06:12.080 --> 00:06:20.440
Now A to the 4B squared is 7 and a ^3 b
^3 is 21 / 2.

1fe70f5b-1761-49b0-9cb9-057f8a911432-0
00:06:20.440 --> 00:06:23.160
So I need to divide by that 21 / 2.

a0d055a0-22d5-467c-8fcf-a36fca0ff578-0
00:06:23.640 --> 00:06:28.000
So that's the same as multiplying by 2 /
21.

55af9a24-9f4f-4cc8-b654-726fa6bb4bdc-0
00:06:28.600 --> 00:06:30.600
So I'll just get rid of that line there.

28981a6e-1dd5-439c-88ac-791f41f1e2bd-0
00:06:31.200 --> 00:06:36.360
So this will be this value we can see
this is the same as 14 / 21.

4b2d8d2a-7826-4bbf-8f15-60c6ebc8e7aa-0
00:06:37.000 --> 00:06:38.640
We can see the factor of seven there.

7fbb4225-738d-4775-8e4d-9cbe3cfe0aa6-0
00:06:38.640 --> 00:06:40.160
This will give us 2/3.

e86c1aba-d7e5-499f-a154-7c4b84f9b0ee-0
00:06:40.480 --> 00:06:45.840
So now we have the a / b must be 2/3 and
we we need to square that.

bf024010-330c-4abb-9178-b63c0229eff3-0
00:06:45.840 --> 00:06:48.320
So squaring 2/3 will give us 4 / 9.

95012110-de84-40f4-8ba6-a19a355c77a0-0
00:06:48.720 --> 00:06:50.920
And we see that over here as answered
this.

3cd9652f-d75d-4e65-9724-d5b24cc62131-0
00:06:52.240 --> 00:06:57.149
OK, so to reflect on this question,
you will have noticed in the solution I

3cd9652f-d75d-4e65-9724-d5b24cc62131-1
00:06:57.149 --> 00:07:00.379
provided,
I calculated the values of 6 choose two

3cd9652f-d75d-4e65-9724-d5b24cc62131-2
00:07:00.379 --> 00:07:03.480
and six choose three pretty quickly in my
head.

b4dde125-0923-44f2-8970-4999622df7e0-0
00:07:03.880 --> 00:07:06.720
So here's a little bit about what was
going on with that.

b29c4b34-d93a-4a82-838b-6db270486d7a-0
00:07:07.400 --> 00:07:11.413
We can recall the formula for N choose K
in general,

b29c4b34-d93a-4a82-838b-6db270486d7a-1
00:07:11.413 --> 00:07:17.320
and that is always N factorial divided by
K factorial times n -, K factorial.

9f8e2f75-12ab-4734-826b-3a5e1a5ca871-0
00:07:18.040 --> 00:07:22.800
So here we see that formula applied to
give 6 choose two i.e.

06693d0a-72cf-4aaf-abd7-9622406c3026-0
00:07:22.880 --> 00:07:24.920
n = 6 and K = 2.

e395d2d3-75f9-4cf6-b975-601da6ce8ef6-0
00:07:25.280 --> 00:07:27.760
So we have 6 factorial as N is 6.

0e9dc9ab-b7c8-4376-8597-b313cfdfbca2-0
00:07:28.080 --> 00:07:33.560
We have two factorial as K is 2 and then
N minus K 6 -, 2 will be 4.

1915c61c-a37e-49fb-8233-0d65dcd011cb-0
00:07:33.560 --> 00:07:35.120
We get this 4 factorial here.

78907f26-2c05-468f-8bd0-953fae21f5df-0
00:07:36.680 --> 00:07:43.321
If we look more closely at this,
we can see that the top is 6 * 5 * 4 * 3

78907f26-2c05-468f-8bd0-953fae21f5df-1
00:07:43.321 --> 00:07:44.040
* 2 * 1.

854ba3e3-a63e-4b00-9308-c7ebeaab85bd-0
00:07:44.240 --> 00:07:50.685
In other words, times 4 factorial,
and this will be over 2 factorial,

854ba3e3-a63e-4b00-9308-c7ebeaab85bd-1
00:07:50.685 --> 00:07:55.565
4 factorial,
and so we can cancel the four factorial

854ba3e3-a63e-4b00-9308-c7ebeaab85bd-2
00:07:55.565 --> 00:08:01.550
from the top and the bottom,
and we are left with 6 * 5 / 2 * 1,

854ba3e3-a63e-4b00-9308-c7ebeaab85bd-3
00:08:01.550 --> 00:08:03.760
which gives us 30 / 215.

6c0417d8-7ef1-45f5-9fe7-785f1d2b6e33-0
00:08:05.680 --> 00:08:10.154
So we can always use this trick to speed
up calculations of these Bernoulli

6c0417d8-7ef1-45f5-9fe7-785f1d2b6e33-1
00:08:10.154 --> 00:08:10.920
coefficients.

cfc2c982-a6f6-40e8-80f0-11dab1f4f666-0
00:08:11.280 --> 00:08:12.040
Let's look at six.

a516843c-9c82-4799-bb22-b55aa1ff28e4-0
00:08:12.040 --> 00:08:12.880
Choose two.

ea5b98ba-eba9-4cac-b465-9c9fa0f6cd07-0
00:08:13.160 --> 00:08:15.720
We see that it's 6 * 5.

27ecf8e9-0016-41a3-b47e-aa6148dd5f8a-0
00:08:16.000 --> 00:08:21.904
So the number of things being multiplied
together on the top is this value here 6

27ecf8e9-0016-41a3-b47e-aa6148dd5f8a-1
00:08:21.904 --> 00:08:26.440
* 5 and then on the denominator we always
have this factorial.

4db1f83a-2e7e-4cfd-b53b-f0284c86a099-0
00:08:26.920 --> 00:08:35.916
So if I apply this to get 6 choose three,
I can say that six choose three would be

4db1f83a-2e7e-4cfd-b53b-f0284c86a099-1
00:08:35.916 --> 00:08:37.000
6 * 5 * 4.

47164e57-6d46-4d64-ba9d-0ea7aea25daa-0
00:08:37.480 --> 00:08:42.175
So I have value starting from six going
down multiplied together,

47164e57-6d46-4d64-ba9d-0ea7aea25daa-1
00:08:42.175 --> 00:08:46.727
and I take three of them,
this value here and then that will be

47164e57-6d46-4d64-ba9d-0ea7aea25daa-2
00:08:46.727 --> 00:08:49.360
divided by three factorial 3 * 2 * 1.

70ecf256-b72a-4fd0-ba22-03fe6b445236-0
00:08:50.280 --> 00:08:54.080
And notice I can spot that I have a six
here over a six.

ce2bdb6c-834d-4f02-b6a1-964089f90813-0
00:08:54.080 --> 00:08:58.920
So these cancel and it leaves me with 5 *
4 giving 20.

374c7f02-c366-4508-a4e5-07be7cae2ed5-0
00:08:59.440 --> 00:09:03.641
So it's well worth practising this sort
of idea to speed up your mental

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arithmetic for binomial coefficients.

678da5b2-41eb-4086-ad93-9f24932e1bb2-0
00:09:05.800 --> 00:09:08.560
This this trick will speed things up a
lot.

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And often you can see some nice
cancellation even after you've used that

129ffe8a-d3c7-4c9b-afaf-8e1b4ed561ca-1
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idea at the start.