00:01
We are going to find the solution to each of these recurrence relations with the given initial conditions.
00:09
For that, we use an iterative forward substitution approach.
00:14
So in part a, we have a -n equal a -n -1 plus n, and a -0 is 4, and in part b, an is 3 -n -minus -1 plus 2, and a -0 is 1.
00:28
So let's start with part a.
00:30
So we are going to substitute progressively for a1, a2 and so on, and we deduce a formula for that from the expressions we are going to get.
00:43
So a1 will be a0 plus 1 because remember here we use the same n.
00:54
Here is the index for the term we are calculated.
00:56
So a1 is the previous one a0 plus the index 1 here.
01:04
But that's a1 and we can now write a2.
01:09
I'm not going to put a0 for the moment.
01:12
I'm going to put it at the end of the deduction formula.
01:16
So a2 is, apply again the formula, is a1 plus the index here is 2.
01:24
But now we can replace a1 here by this expression.
01:29
So we get a0 plus 1 plus 2.
01:34
Let's do one more.
01:37
So we have a3 by the recursive formula here is a2 plus the index 3.
01:46
But now we can replace a2 by expression here.
01:52
A2 is a0 plus 1 plus 2.
01:54
So we get a0 plus 1 plus 2.
01:58
That's a 2 plus 3.
01:59
So it's not hard to verify that in general, a n will be a 0 plus the sum of the positive integers from 1 to the index we are calculated.
02:21
That is 1 plus 2 plus up 2.
02:28
When we know a formula of this sum, we have a formula for that, and that formula is the last term in the sum n times the next one.
02:40
And plus 1 over 2 so the recursive relation can be written in an explicit way as a n equal a 0 plus n times n plus 1 over 2 and now we use a 0 is 4 so the solution here is a n equal 4 plus it's 4 plus n times n plus 1 over 2 that for n greater equal to 2 sorry created down or equal to 1 in this case we we're going to verify at for 0 we get also the result so we include zero value in the indices so this is expression we were looking for and now we verify well verify some terms.
04:00
So a zero for example if we put zero here we get zero times one for two is zero this bar here and so we get four and put it here zero times one over two is four that's correct.
04:15
A1 is four plus one times two over two so it's four plus one with five.
04:33
A2 is four plus two times three over two is seven let's do one more a three is four plus three times four over two four over two is two times three six so four plus six is ten okay let's verify now using the recursive formula a zero is four is given a one is a zero plus one that is four plus 1 equal 5.
05:15
It's the same we got over here.
05:19
A2 is a1 plus the index 2.
05:24
That is a1.
05:25
We calculated here is 5 plus 2 is 7.
05:28
And we got the same as here.
05:31
And a3, for example, is a2 plus the index 3 is 7 plus 3, is 10, and so on.
05:42
So we have found the general formula or the explicit formula if you want for.
05:49
Let me get rid of this for a moment.
05:52
We found the split formula for the recurrence relation in part a.
05:56
We go to part b now.
06:02
And in part b we have an equals 3an minus 1 plus 2.
06:12
And the first term a0 is 1.
06:17
Good...