1 00:00:01,069 --> 00:00:04,370 Welcome to a new lecture on the binomial theorem. 2 00:00:04,370 --> 00:00:08,781 In this lecture, we will look at certain divisibility results which can be derived based on the 3 00:00:08,781 --> 00:00:10,950 binomial theorem. 4 00:00:10,950 --> 00:00:20,460 Once again, recall the binomial theorem is given by (x+y)^n = \sum{r=0}^{n}{C_r x^{n-r} 5 00:00:20,460 --> 00:00:27,400 y^r}, where we use C_r to denote nC_r. 6 00:00:27,400 --> 00:00:31,630 To give you a flavor of the kind of results we will be looking at today, let's look at 7 00:00:31,630 --> 00:00:56,770 the first example which asks you to show that 3^{2n+2} - 8n - 9 is divisible by 64 when 8 00:00:56,770 --> 00:00:59,420 n is a natural number. 9 00:00:59,420 --> 00:01:10,560 It asks you to show that this term is divisible by 64 for natural numbers n. 10 00:01:10,560 --> 00:01:17,530 At first sight, there doesn't seem to be a binomial expansion you can apply to generate 11 00:01:17,530 --> 00:01:25,440 this result so we would like to create a binomial expansion to derive this result. 12 00:01:25,440 --> 00:01:31,580 Notice that you have an exponent here, but you also have some terms here and these terms 13 00:01:31,580 --> 00:01:36,408 are, in general, not divisible by 64 for n being a natural number. 14 00:01:36,408 --> 00:01:40,800 You can plug in certain natural numbers and you can see that this term, which is being 15 00:01:40,800 --> 00:01:47,030 subtracted from 3^{2n+2}, is not going to be divisible by 64 and therefore we would 16 00:01:47,030 --> 00:01:52,170 like to somehow get rid of these terms and leave the remaining term to be divisible by 17 00:01:52,170 --> 00:01:54,280 64. 18 00:01:54,280 --> 00:01:56,840 Let's look at a way of re-expressing this term. 19 00:01:56,840 --> 00:02:10,649 3^{2n+2} - 8n - 9 can be written as 9^{n+1}, where I've just taken a power of 2 into 3 20 00:02:10,649 --> 00:02:18,579 and written 3^2 as 9, and you still have - 8n - 9. 21 00:02:18,579 --> 00:02:23,550 So now, this looks a little more promising because you have 9^{n+1} and you have a 9 22 00:02:23,550 --> 00:02:24,550 and you have an 8n. 23 00:02:24,550 --> 00:02:35,730 You can get rid of the 8n potentially by writing 9 as (1+8)^{n+1} and you have - 8n - 9. 24 00:02:35,730 --> 00:02:40,030 So this looks a little more promising, and now you have a term which you can expand using 25 00:02:40,030 --> 00:02:43,410 the binomial theorem. 26 00:02:43,410 --> 00:02:45,390 Let's try and expand this term. 27 00:02:45,390 --> 00:03:19,210 This is nothing but (n+1)C_0 1^{n+1} + (n+1)C_1 1^n 8 + (n+1)C_2 1^{n-1} 8^2 + .... We'll 28 00:03:19,210 --> 00:03:26,160 be concerned about the higher-order terms later, and we have/we're still subtracting 29 00:03:26,160 --> 00:03:30,030 8n and 9 from this term. 30 00:03:30,030 --> 00:03:32,290 Let's see what this simplifies to. 31 00:03:32,290 --> 00:03:43,370 This is nothing but: the first term is 1; the second term is (n+1)*1*8; and the remaining 32 00:03:43,370 --> 00:04:04,280 terms can be succinctly written as summation \sum_{r=2}^{n+1}{(n+1)C_r 1^{n+1-r} 8^r}. 33 00:04:04,280 --> 00:04:09,020 You still have to subtract 8n and 9 from this expression. 34 00:04:09,020 --> 00:04:15,030 This looks a little more promising, because notice that these first two terms add up to 35 00:04:15,030 --> 00:04:20,320 exactly 8n+9 and you can cancel these two terms. 36 00:04:20,320 --> 00:04:35,590 So, you're just left with summation \sum_{r=2}^{n+1}{(n+1)C_r 1^{n+1-r} 8^r}. 37 00:04:35,590 --> 00:04:40,860 This is what we're left with, and we want to show that this is divisible by 64. 38 00:04:40,860 --> 00:05:03,000 Notice that this summation runs from r=2, so let's rewrite this as \sum_{r=0}^{n-1}{(n+1)C_{r+2}8^{r+2}}. 39 00:05:03,500 --> 00:05:09,940 We've just used a change of indices to rewrite this summation. 40 00:05:09,940 --> 00:05:13,140 Notice that there is a 8^2 here; you can pull that out. 41 00:05:13,140 --> 00:05:26,010 This is going to be 64*[\sum_{r=0}^{n-1}{(n+1)C_{r+2} 8^r}]. 42 00:05:26,010 --> 00:05:30,350 So you have a 64 here, and you wanted to show that this term is divisible by 64: so that 43 00:05:30,350 --> 00:05:31,410 is good. 44 00:05:31,410 --> 00:05:38,960 What we have to show is that this term is an integer, that's all is what remains. 45 00:05:38,960 --> 00:05:45,700 That term is clearly an integer because you have a summation of several terms, where each 46 00:05:45,700 --> 00:05:49,820 term is made up of two integers themselves. 47 00:05:49,820 --> 00:05:53,870 8^r is an integer and (n+1)C_{r+2}, again, is an integer. 48 00:05:53,870 --> 00:05:57,409 We have summations of integers and so this term is an integer. 49 00:05:57,409 --> 00:06:05,380 So we have that this overall term is divisible by 64, which is the desired result. 50 00:06:05,380 --> 00:06:12,330 We've seen an application of the binomial theorem where we somehow re-expressed our 51 00:06:12,330 --> 00:06:20,750 given term in terms of an expression from where we could use the binomial expansion 52 00:06:20,750 --> 00:06:25,460 and derive a divisibility result for the original term. 53 00:06:25,460 --> 00:06:31,110 Let's look at a second example. 54 00:06:31,110 --> 00:06:56,860 This example asks to show that 2^{4n} - (2^n)*(7n+1) is divisible by 196 whenever n is a natural 55 00:06:56,860 --> 00:07:00,080 number. 56 00:07:00,080 --> 00:07:05,240 This example is very similar to the previous one, and once again we have to figure out 57 00:07:05,240 --> 00:07:11,660 how to rewrite this original term in a form which is amenable to the binomial expansion. 58 00:07:11,660 --> 00:07:19,000 Notice, now, that we want to show that this term is divisible by 196, and 196 is nothing 59 00:07:19,000 --> 00:07:27,510 but (14)^2 and you have a 7 here and a 2^n here, so that looks promising. 60 00:07:27,510 --> 00:07:32,430 Let's see what we can do. 61 00:07:32,430 --> 00:07:43,100 2^{4n} - (2^n)*(7n+1) can be rewritten as (16)^n - (2^n)*(7n+1), where we've just rewritten 62 00:07:43,100 --> 00:07:46,380 2^{4n} as (16)^n. 63 00:07:46,380 --> 00:08:00,760 Now, 16 can somehow be split into 14 and 2; so, we can rewrite this as (2+14)^n - (2^n)*(7n+1). 64 00:08:00,760 --> 00:08:07,770 This looks promising because, now, we have a 14 appear in the binomial term and we want 65 00:08:07,770 --> 00:08:14,380 to show that this whole term is divisible by 196 which is (14)^2. 66 00:08:14,380 --> 00:08:20,400 Let's go through the motions again and use the binomial theorem to show that this is 67 00:08:20,400 --> 00:08:34,830 nothing but nC_0 2^n (14)^0 + nC_1 2^{n-1} (14)^1 + (let's rewrite the higher order terms 68 00:08:34,830 --> 00:08:48,990 as) summation \sum_{r=2}^{n}{nC_r 2^{n-r} (14)^r} ,and we subtract the original (2^n)*(7n+1) 69 00:08:48,990 --> 00:08:52,400 from this expansion. 70 00:08:52,400 --> 00:09:03,490 Once again, you can rewrite this as 2^n + n*(2^{n-1})*14 + \sum_{r=2}^{n}{nC_r 2^{n-r} 71 00:09:03,490 --> 00:09:18,610 (14)^r} and you still have to subtract (2^n)*(7n+1). 72 00:09:18,610 --> 00:09:33,510 This is nothing but 2^n + n*(2^n)*7, where I've factorized 14 as 2*7, and you have a 73 00:09:33,510 --> 00:09:48,690 resulting summation \sum{r=2}^{n}{nC_r 2^{n-r} (14)^r}, and you subtract (2^n)*7n and a 2^n, 74 00:09:48,690 --> 00:09:52,880 where I've just expanded this term. 75 00:09:52,880 --> 00:09:58,070 Notice once again that you can cancel the first two terms with the terms being subtracted, 76 00:09:58,070 --> 00:10:04,160 and you're still left with/you're only left with the summation \sum_{r=2}^{n}{nC_r 2^{n-r} 77 00:10:04,160 --> 00:10:05,560 (14)^r}. 78 00:10:05,560 --> 00:10:11,640 Now the arguments are quite similar to the previous one/previous example we've seen, 79 00:10:11,640 --> 00:10:13,000 so I won't go through this in detail. 80 00:10:13,000 --> 00:10:49,810 You can just rewrite this as (14^2)*[\sum_{r=0}^{n-2}{nC_{r+2} 2^{n-r-2} (14)^r}], where here you started 81 00:10:49,810 --> 00:10:58,350 with 2^{n-2} so you're starting with 2^{n-2} again; and here you started with (14)^2 whereas 82 00:10:58,350 --> 00:11:05,620 here you start with (14)^0 since I've pulled out a (14)^2 here. 83 00:11:05,620 --> 00:11:10,220 This is what we're left with, and notice that this summation again is an integer and we 84 00:11:10,220 --> 00:11:16,089 have this summation being multiplied by (14)^2 which is 196 and which is again an integer. 85 00:11:16,089 --> 00:11:20,810 We wanted to show that this term is divisible by 196, and that's exactly what we've shown 86 00:11:20,810 --> 00:11:27,180 since we've factorized it into a form where we have a leading coefficient of (14)^2 and 87 00:11:27,180 --> 00:11:32,690 so, we've once again shown that this term is divisible by 196 whenever n is a natural 88 00:11:32,690 --> 00:11:37,190 number. 89 00:11:37,190 --> 00:11:46,180 Let's look at a multiple choice question which is based on this flavor of problems. 90 00:11:46,180 --> 00:12:10,950 The question asks you to show/asks you when 5^{99} is divided by 13, the remainder is: 91 00:12:10,950 --> 00:12:15,670 it asks you to compute the remainder when 5^99 is divided by 13. 92 00:12:15,670 --> 00:12:31,490 The options are 8, 9, 10, and "none of these". 93 00:12:31,490 --> 00:12:36,020 This is a kind of problem where it's hard to eliminate the options without actually 94 00:12:36,020 --> 00:12:42,230 solving the problem because you have "none of these". 95 00:12:42,230 --> 00:12:46,580 Even if you determine that this remainder is going to be even or odd, you are not really 96 00:12:46,580 --> 00:12:52,050 sure which option it is since there's a "none of these" and, even if the remainder is odd, 97 00:12:52,050 --> 00:12:55,080 you can't just conclude that the answer is 9. 98 00:12:55,080 --> 00:12:59,959 So you would actually have to derive the result for this example. 99 00:12:59,959 --> 00:13:06,110 Let's see how we can rewrite 5^{99} to obtain an amenable result. 100 00:13:06,110 --> 00:13:15,720 To arrive at the solution, let's try computing some powers of 5 to see where that gets us. 101 00:13:15,720 --> 00:13:23,740 5^1 is 5, which is quite far from 13: there's not really much you can do about it. 102 00:13:23,740 --> 00:13:33,940 5^2 is 25, which is close to a multiple of 13 (13*2 is 26, so this looks promising). 103 00:13:33,940 --> 00:13:43,470 But 5^{99} is not a direct power of/not an integral power of 5^2, and so let's try this 104 00:13:43,470 --> 00:13:52,260 approach: let's write 5^{99} as 5*(5^{98}) where I've just factorized 5^{99} into a more 105 00:13:52,260 --> 00:13:53,790 suitable form. 106 00:13:53,790 --> 00:13:58,010 Now I can write this as 5*((5^2)^{49}). 107 00:13:58,010 --> 00:14:11,250 Since I know that 5^2 is close to a multiple of 13, this is nothing but 5*(26 - 1)^{49}. 108 00:14:11,250 --> 00:14:26,680 26 is a multiple of 13, so the divisibility results are going to be fairly easy to derive. 109 00:14:26,680 --> 00:14:35,240 This is nothing but 5*..., and now you use the binomial expansion of (26-1)^{49}. 110 00:14:35,240 --> 00:14:43,100 Let's see how we can write this. 111 00:14:43,100 --> 00:14:53,190 5*[49C_0 (-1)^{49} (26)^0 (I'm starting off with (-1)^{49} and (26)^0 since we're only 112 00:14:53,190 --> 00:14:58,550 concerned with the power of 26 which is 0, since all other terms are going to be divisible 113 00:14:58,550 --> 00:15:15,430 by 13) and you have + 49C_1 (-1)^{48} (26)^1 + ...]. This is going to be the expansion 114 00:15:15,430 --> 00:15:22,390 of (26-1)^{49}, and you have a leading term which is 5 in this expression. 115 00:15:22,390 --> 00:15:25,790 Notice that all of these higher powers, as I mentioned, are going to be divisible by 116 00:15:25,790 --> 00:15:32,200 13, and, since we only care about the remainder that this term leaves when divided by 13, 117 00:15:32,200 --> 00:15:37,670 we don't really care about these terms which are exactly divisible by 13. 118 00:15:37,670 --> 00:15:43,970 This is the only remaining term, which is going to give us the remainder when you divide 119 00:15:43,970 --> 00:15:48,230 5^{99} by 13 so let's see what that gives us. 120 00:15:48,230 --> 00:15:56,750 5*[49C_0, which is 1, times (-1)^{49}, which is -1, times (26)^0, which is 1], and so this 121 00:15:56,750 --> 00:15:58,570 gives us -5. 122 00:15:58,570 --> 00:16:04,839 Therefore, the remainder when 5^{99} is divided by 13 is -5. 123 00:16:04,839 --> 00:16:09,420 For those of you who aren't used to this kind of notation, this might seem a little strange 124 00:16:09,420 --> 00:16:15,260 that you get a negative remainder but, to get a positive remainder, you just add a multiple 125 00:16:15,260 --> 00:16:16,260 of 13. 126 00:16:16,260 --> 00:16:20,230 Since we've ignored so many multiples of 13, you get a negative number but, if you add 127 00:16:20,230 --> 00:16:24,340 a multiple of 13, you get 8. 128 00:16:24,340 --> 00:16:29,709 The remainder when 5^{99} is divided by 13 is 8. 129 00:16:29,709 --> 00:16:35,120 For those of you who are familiar with modular arithmetic, notice that the approach we took 130 00:16:35,120 --> 00:16:40,290 is very similar to the approach you take when you use/when you do such kinds of problems 131 00:16:40,290 --> 00:16:48,230 in number theory where you try to get at a power of 5 which is close to a multiple of 132 00:16:48,230 --> 00:16:56,240 13 and you just use remainder arguments, where you ignore terms which were multiples of 13, 133 00:16:56,240 --> 00:17:01,550 and so you're only left with one term since you only care about the remainder. 134 00:17:01,550 --> 00:17:07,169 This is a sample of a problem which could appear in an exam-type situation where you 135 00:17:07,169 --> 00:17:15,269 are asked to derive a result based on the divisibility of a term by a natural number 136 00:17:15,269 --> 00:17:17,789 using the binomial theorem. 137 00:17:17,789 --> 00:17:19,160 That's it for this lecture. 138 00:17:19,160 --> 00:17:21,659 Hope you enjoyed watching this lecture. 139 00:17:21,659 --> 00:17:27,319 In the next lectures, we will look at some example problems based on the binomial theorem. 140 00:17:27,319 --> 00:17:31,210 I thank you for your attention, and hope you can join us next time.