00:01
So, you have given here a sequence that is yn is equal to 3 into minus 1 to the power n and where n is equal to 0 ,1 ,2 ,3 and up to 11.
00:20
Now we can write our y0n that is equal to into two terms.
00:28
It gives 3 value when n is equal to even number means 0 ,2 ,4 and up to we have here 10 and it gives minus 3 value if n is odd number n is equal to 1 ,3 ,4 ,5 ,7 ,9 and 11.
00:59
So these are the condition.
01:00
Now we can write the above calculation as yk is equal to summation of n is equal to 0, k, yn, e to the power minus of iota pi by 6, nk.
01:17
So according to the above equation we can write y0 is equal to we have y0 plus y2 plus y4 plus y6 because we need to find here summation and plus y6 here is the y6 and here is the y8 and plus y10 and second sequence we have given here the odd number.
01:49
So we have here y1 plus y3 plus y5 plus y7 plus y9 plus y11.
02:04
So after putting here value we have 3 plus 3 plus 3 plus 3 because in even number function gives 3 value.
02:16
So we have to put here all even number value is 3 plus 3 plus 3.
02:23
So we can write here and plus minus 3 minus 3 minus 3 minus 3 and here is the minus 3 because these all are odd number.
02:37
So it gives us 0 value because positive number cancelled by negative number.
02:42
Now we need to find here y1.
02:44
So y1 is equal to then we have y0 plus y2 e to the power minus of iota pi by 6 nk multiply by 2 plus y4 e to the power minus iota upon pi by 6 multiply by 4 plus and up to we can write this y10 e to the power minus iota pi by 6 multiply by 10.
03:21
So these all are even number.
03:23
Now odd number are that is plus y1 and e to the power minus iota pi by 6 and similarly up to we can write this these all number up to 11.
03:34
So we can write here y11 here is the plus here is the plus e to the power minus iota pi by 6 multiply by 11...