00:01
For this problem, we have been given eight recursive relationships with initial conditions for different sequences.
00:08
Now, the recursion method is a really nice, compact way to put down sequences.
00:15
Unfortunately, to find like the 100th sequence, or the 100th term in the sequence requires us to know all of the other previous 99.
00:23
So sometimes it's nice to be able to write these sequences without having to use the recursive.
00:30
Elements.
00:31
So we're going to take recursive definitions of sequences and change them so that they are not recursive.
00:38
So let's take a look at the first one and see how this works.
00:42
We have a to the n equals three times a to the n minus one.
00:48
So that's our recursive definition.
00:50
And we're also told that a sub zero equals two.
00:54
So what we're going to do for this problem, and all of the subsequent ones in this example, we're going to find maybe the first, maybe up through a sub 4, and see if we can find a pattern that emerges, that we could use to define this sequence.
01:10
A sub 1 means i take three times my prior one.
01:14
So that'll be three times two.
01:16
How about a sub 2? well, that's going to be three times the prior term, three times two.
01:23
A sub 3 will be three times the prior term, which is three times three times two.
01:32
And you can hopefully see the pattern emerging here.
01:35
In each case, i have three to the same power as the base.
01:43
A sub 4 has three to the fourth.
01:45
A sub three has three to the third.
01:48
So we can take the sequence and say the nth term of my sequence will be three raised to that nth power times 2.
02:00
So while the recursive definition is nice, our new definition, we can find the thousandth term and only the thousandth term and not have to find all of the ones leading up to it.
02:10
So it is convenient to be able to rewrite these.
02:14
Okay, b, our recursive definition says a sub n equals a sub n minus 1 plus 2.
02:23
And we're going to start a sub 0 equals 3.
02:27
Well, again, let's do the first few here and see if we can see a pattern emerge.
02:35
Okay, so a sub 1, we take the prior term plus 2.
02:41
A sub 2, we take the prior term plus 2.
02:47
A sub 3, we take the prior term plus 2.
02:52
Sometimes it's useful to actually do the math here, to do the addition or the multiplication.
02:58
If possible, i'm going to leave it out, spread out like this, because rather than just seeing three and five and seven and nine, this really shows how the numbers are getting put together.
03:07
And i think it makes it a little easier to see the pattern.
03:11
Okay, let's just do one more.
03:13
That's the prior term plus two.
03:17
So in this case, if i look at what the pattern is, in each case, i have a three in the beginning.
03:24
So three plus.
03:26
And then i have, i'm adding two, and i'm adding one more two each term.
03:32
So i'm adding two times n.
03:36
So for the a sub three, i've got three twos.
03:39
For a sub four, i have four twos.
03:42
So here is my definition for this sequence.
03:47
Next, i have a sub n equals a sub n minus 1 plus n.
03:57
And in this case, we are starting with a sub 0 equals 1.
04:02
So let's put out a few terms here.
04:07
Okay, a sub 1, i take the prior term.
04:09
Plus n.
04:11
Okay.
04:12
A sub two, i take the prior term plus n.
04:17
Prior term plus n.
04:22
Prior term plus n.
04:26
Okay, so what pattern do we see emerging here? well, in each case, it's pyramitting.
04:34
So i have, i have this plus one just in the front here.
04:39
So i think whatever i do, i'm going to say one plus.
04:41
And then i've got a pyramid.
04:43
I've got a sub 1 is 1.
04:45
A sub 2 is 1 plus 2.
04:47
A sub 3, 1, 2, 3, 1, 2, 3, 1, 2, 4.
04:51
Well, we do have a formula for adding the first n integers.
04:56
It is n times n plus 1 divided by 2.
05:02
If you recall why this works, i'll take a sub 4 as my example.
05:07
I have, if i add the first and the last, i have, that's n and then plus 1.
05:15
So that's 4 and 1 is 5.
05:18
And every pairing is going to equal that same amount, and i'm going to have n over 2 of those pairs.
05:24
So n plus 1 times n over 2 pairs.
05:29
So this is my sequence, or by the definition for this sequence, not using a recursive definition.
05:39
Okay, on to d.
05:43
D says a sub n equals a sub n plus 2n plus 2n plus.
05:51
3.
05:52
And i'm going to start with a sub 0 equaling 4.
05:58
As 2, a sub 3, a sub 4.
06:01
Okay, so let's see what we get.
06:04
First, a sub 1, we take the prior term plus 2 times 1 plus 3.
06:12
A, a sub 2, we take the prior term plus 2 times n plus 3.
06:23
A sub 3, prior term, plus, well, that 2 times 2, less, well, that 2 times 2, less just kind of put that together here, plus two times three, plus three.
06:36
Let's just do one more.
06:38
Unfortunately, i'm not really seeing a solid pattern come out.
06:43
There's the prior term plus two times n plus three.
06:49
Lots of fours and twos and threes, but nothing really is jumping out at me.
06:54
So let's try another tack.
06:56
Let's actually add them up and see what we get.
06:59
Okay, and i'm just change a different color here.
07:01
So i've got four for the first.
07:02
One.
07:03
Next i have nine.
07:06
Then i have 16.
07:10
Okay, now i see a pattern.
07:13
By adding these, i can see that i'm going up.
07:15
These are all perfect squares.
07:17
And i'll just put this one in blue as well so you can really see that pattern.
07:21
Now, this is not n squared though.
07:25
But when n is zero, it's two squared.
07:28
When n is one, it's three squared.
07:30
So it looks like i'm taking two plus n.
07:33
And that's what i'm squaring.
07:36
So so when a sub n is 4, it's n plus 2 or 6 squared.
07:43
When n is 3, it's 2 more than that, or 5 squared.
07:47
So there's that one.
07:48
So sometimes it's easier to see it when you write it all out.
07:51
Sometimes adding it shows the pattern best.
07:55
Let's do e.
07:56
A sub n equals 2 times a sub n minus 1, minus 1.
08:05
And we're going to start with a sub 0 equaling 1.
08:12
Okay, so let's begin.
08:14
A sub 1, i take 2 times the prior 1 minus 1.
08:22
That equals 1.
08:24
That's exactly where i just started.
08:26
A sub 2, i'll take 2 times the prior 1 minus 1 equals 1...