مجموعهای از دنباله ها

سرفصل: بخش ریاضی / سرفصل: مسائل کلمه / درس 20

مجموعهای از دنباله ها

توضیح مختصر

  • زمان مطالعه 5 دقیقه
  • سطح متوسط

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Sums of Sequences

Our final topic in sequences concerns the Sum of Sequences. Sometimes the test will ask us for the sum of a long sequence. For example what is the sum of all the multiples of 20 from 160 to 840? So obviously, that’s far too long to add up the individual numbers, so we need some kinda shortcut. A famous legend from the history of mathematics will help us figure out this sort of question.

So I’ll tell you this story first of all. The great mathematician Carl Friedrich Gauss, known as the Prince of Mathematics, was precocious as a, as a child. And so there are a few different versions of this story, but one version of this story is, he was five years old and he was pestering his kindergarten teacher with too many questions, and she wanted to kind of shut him up and keep him busy with something for awhile.

She asked him to add all the integers from 1 to 100, and to her astonishment, Gauss produced the answer in seconds. How did he do that? The five-year old Gauss noticed the following. Suppose we have this sum, all the integers from 1 to 100. Well, take out the sum of the first and the last.

Just take out the 1 and the 100. Of course, they add up to 101. Well now take out the next. Once those first two are missing, what’s left is the first and the last. The 2 and the 99. Take that out also.

Two and 99 also add up to 101. Then the next pair, 3 and 98. They also add up to 101. Been four and 97 and five and 96 and so forth. Clearly we continue to pair all the numbers in the sequence this way, we would wind of with 50 pairs, each of which has a sum of 101.

Well that makes it very easy. That means that the total sum is 50 times 101, or another words 5,050. So that’s actually a very easy sum to do once you see this trick. Let’s generalize that to the sum of the first N positive integers. So we have all the integers 1, 2, 3, 4 dot, dot, dot all the way up to N. Well much in the same way, we’re gonna pair in the same way from the outer ends, and so each pair is going to add up to N plus 1.

We take the 1 and the N that adds up to 1 plus N plus 1. If we take the N minus 1 in the 2 that adds up to N plus 1, and in fact all the terms when we pair them this way add up to the same sum, N plus one, and of course there are N over two pairs, so the sum of the entire sequence has to be N over 2 times N plus 1 and we get that formula at the bottom. Now, it might bother you a little bit, what if N is odd?

Then there’s, then N over 2 is not gonna be an integer, it’s gonna be something half. And that’s many, how many pairs we have. Well, keep in mind that even if N is odd, N plus 1 is even. And so in fact this all works out. This formula works regardless of whether N is even or odd.

More generally, suppose we have any evenly spaced list that has N items with a lowest term a sub 1, and a final term a sub N. If we pair from the outside, each pair will add up to a sub 1 plus a sub N, which is the sum of the first and the last terms. If there are N total terms, there will be N divided by 2 pairs of terms and each pair will have the same sum.

Thus, the sum of the list will be given by that formula. So this is the sum of the list of any evenly spaced list of numbers. And notice we can also think of this as N, the number of items on the list, times the average of the first and last pair. That’s another perfectly valid way to think about this formula. So here’s a practice question, the same practice question that appeared on the first slide.

Pause the video and see if you can solve this now. Okay, what is the sum of all the multiples of 20 from 160 to 840 inclusive? First, we have to know how many terms we have here. So, 160 is 8 times 20, so it is the eight multiple. 840 is 42 times 20, so that is the 42nd multiple of 20. And inclusive counting, we would do 42 minus 8 plus 1 that’s 35.

Number of pairs divide that by 2 we get 17.5, that will not be a problem. And the sum of the list will be the number of pairs times the sum of the first and the last term. Well very conveniently the first and the last term add up to 1000, so we just multiply and we get 17,500. To add sequences of evenly spaced numbers, pair the numbers by first and last to produce a pair, pairs of a constant sums, then multiply by the number of pairs.

Once again I would urge you not merely to memorize the formulas here, instead understand the logic of the procedure explained. If you remember the story, if you remember the whole procedure, then you will understand the formula much more deeply.

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