00:01
Hi, we're giving here h1, h1, s3 and so until h2n.
00:04
Our 2 in harmony means between a and d.
00:07
I'm going to solve for this expression here.
00:10
So we'll have 1 over a, 1 over h1, 1 over h2, and so on 1 over h2, and so on 1 over h2n.
00:20
We'll be an ap, alright? that will be an ap, so we can say that last term.
00:27
1 over b equals 1 over a plus 4 number of times 2 n positive 2, negative 2, negative 1, times so we have d from here we calculate d selling out to be a negative b over ab 2 n plus 1.
00:44
So we go to the value of d now i find out we'll find out x1 and h2n so find out here x1 we'll have 1 over h1 equals 1 over a plus d a negative b over ab 2n plus 1 that'll give ab b 2 n plus 1 and then we have b 2n plus 1 positive a negative d that will be 2 and b positive a over ab 2 n plus 1 still go next we'll find out for h2m so 1 over h 2n that equals uh we'll have 1 over b last term negative d a negative b over ab 2 n plus 1 so next coming up to be delta am ab 2 n plus 1 so we get now a 2n plus 1 negative a positive b actually give 2n a positive b over ab 2n plus 1 this we have got it 1 over x 2n 2n now we'll solve for h1 positive a x1 negative a and this so let's work on this.
02:22
First, we solve for x1 in the portion it is given that x1 positive a over h1 negative a...