Would you like to inspect the original subtitles? These are the user uploaded subtitles that are being translated: 1 00:00:00,660 --> 00:00:04,430 Today we're going to be talking about how to list the first several terms of a sequence. 2 00:00:04,500 --> 00:00:10,350 And in this particular problem we've been asked to list the first five terms given that a sub 1 is equal 3 00:00:10,350 --> 00:00:17,340 to one and a formula here for a sub and plus one at the end plus one term which we've been given as 4 00:00:17,410 --> 00:00:20,280 5 a sub n minus 3. 5 00:00:20,390 --> 00:00:24,990 So it a problem like this when we're listing the first several terms the first thing we want to do is 6 00:00:24,990 --> 00:00:28,110 start constructing a table of values. 7 00:00:28,200 --> 00:00:32,730 We know here that our first term a sub one is equal to one so let's go ahead and write that down a sub 8 00:00:32,730 --> 00:00:34,980 1 is equal to 1. 9 00:00:34,980 --> 00:00:39,450 Now we need to find the next term in the sequence Well that's going to be a sub 2 and we don't know 10 00:00:39,450 --> 00:00:43,180 the value of it yet but we know that it's going to be one greater than a sub one. 11 00:00:43,200 --> 00:00:44,970 So we've got here a sub 2. 12 00:00:45,050 --> 00:00:47,270 Okay so how do we find a sub 2. 13 00:00:47,430 --> 00:00:52,000 We're going to be using the formula for a sub and plus one. 14 00:00:52,020 --> 00:00:58,210 Well if we're using this formula here for Aisyah then plus one we know since this is plus one. 15 00:00:58,380 --> 00:01:05,070 And this is n that the value of a sub and plus one is dependent on the previous value. 16 00:01:05,100 --> 00:01:05,900 Subban right. 17 00:01:05,910 --> 00:01:11,290 Because A-7 is exactly one less than a sub and plus one. 18 00:01:11,310 --> 00:01:16,540 So a 7 is going to be US plugging in the previous value that we found. 19 00:01:16,590 --> 00:01:22,500 So to find a sub 2 we're going to get five right which comes directly from our ace and plus one formula 20 00:01:22,980 --> 00:01:25,270 times a sob in the previous term. 21 00:01:25,290 --> 00:01:31,560 Thats a sub once we multiply that by one there because the value of a sub one is one and then we go 22 00:01:31,560 --> 00:01:33,470 ahead and subtract three. 23 00:01:33,660 --> 00:01:40,140 So what we get there then is a sub 2 is equal to 2 5 minus 3 which is 2. 24 00:01:40,170 --> 00:01:48,300 So now we have a value for a sub to a sub 3 the next value is going to involve the exact same formula 25 00:01:48,340 --> 00:01:54,660 but now are going to be using a sub to the previous term before a sub 3s are going to get five times 26 00:01:54,750 --> 00:01:57,770 our previous value here for a sub to 2. 27 00:01:57,960 --> 00:02:01,630 And then we're going to subtract 3 there that's going to give us 10 minus 3. 28 00:02:01,770 --> 00:02:08,910 And we know that a sub 3 is equal to 7 and we just keep going like this until we have the first five 29 00:02:08,910 --> 00:02:16,180 terms of our sequence so a sub 4 will be 5 times the previous value 7 minus 3. 30 00:02:16,260 --> 00:02:25,470 That will give us 35 minus 3 or a sub 4 is equal to thirty two and finally a sub 5 the fifth value or 31 00:02:25,470 --> 00:02:33,570 the fifth term in our sequence will give us 5 times 30 to minus three which is going to give us 160 32 00:02:33,840 --> 00:02:41,670 minus 3 or subtype is equal to one fifty seven and we could just continue on like that for however many 33 00:02:41,670 --> 00:02:44,230 terms we were asked to list. 34 00:02:44,250 --> 00:02:50,250 So thats how you use the first time in a sequence and a formula for the end plus 1 term to list the 35 00:02:50,250 --> 00:02:52,040 first several terms in the sequence. 3968