Find the first six terms of the sequence defined by each of...

Find the first six terms of the sequence defined by each of these recurrence relations and initial conditions. a) an = 6an-1, a0 = 2 b) an = a2n-1, a1 = 2 c) an = an-1 + 3an-2, a0 = 1, a1 = 2 d) an = nan-1 + n^2an-2, a0 = 1, a1 = 1 e) an = an-1 + an-3, a0 = 1, a1 = 2, a2 = 0

Transcript

00:01 So in this problem, we're asked to find the first six terms in these sequences given by each of these recurrence relationships and initial conditions.

00:12 So the first one says ace of n is 6 times aase of n minus 1, where a .m 0 is 2.

00:21 So if n is 1, right, this is n is 1, so i'm at ace of 1, then i have 6 times ace of 1.

00:33 Minus 1, which is 6 times ace of 0, which is 6 times 2, which is 12.

00:43 Okay.

00:44 So the next one.

00:45 Now, n is 2.

00:46 So i have 6 of ace of 2 minus 1, which is 6 times ace of 1, which is going to be ace of 1 is 12.

00:55 So that's 6 times 12, is 72.

01:00 And we keep going.

01:01 6 times ace of 3 minus 1.

01:04 That's 6 of ace of 2.

01:07 Lace of 2 is 72, so this is 6 times 72, which is 432.

01:18 6 times 7 is 42, and 6 times 2 is 12.

01:22 So you'd have 432.

01:25 All right, so we just keep multiplying by 6 each time, don't we? so this is 6 times 432, which is 6 times 2 is 12, carry the 1, 6 times 3 is 18, and 1 is 19.

01:39 Carry the one again 24 that's 25 okay so this is six times 2592 which is going to be 2592 times six that's going to be 15 ,552 and then lastly six times 15 ,552 and then lastly six times 15 ,552 it's going to be 93 ,312 okay next one the relationship is ace of n is 2 times ace of n minus 1, where ace of 1 is 2.

02:26 So i already have ace of 1 is 2.

02:28 So ace of 2 will be 2 times ace of 1, which is going to be 2 times ace of 1, which is 2 times 2, which is 4.

02:41 The next one, 2 times ace of 3 minus 1.

02:45 So that's 2 times ace of 2.

02:47 So that's 2 times 4 which is 8 because a.

02:54 2 we just calculated as 4.

02:56 All right.

02:57 2 times ace of 4 minus 1.

02:59 That's 2 times ace of 3.

03:02 So that's 2 times 8, which is 16.

03:08 So what are we doing each time? we're just multiplying the previous answer by 2, aren't we? so that's going to be 2 times 16, which is 32, and 2 times 32, which is 64.

03:22 Okay.

03:26 So really this recurrence relationship was 2 to the n, wasn't it? it's just squaring, just taking 2 to whatever power n is.

03:39 Okay? let's look at this one now.

03:41 A .s.

03:42 Of n is a .m.

03:43 Minus 1 plus 3 times a .m.

03:46 Minus 2.

03:47 With a .0 is 1, a sub 1 is 2.

03:50 So i have the first 2 already, right? i'm only looking for the first 6.

03:54 So i started at 8 .0.

03:56 It means i can quit at ace of 5.

03:58 So a.