https://youtu.be/Y11HGQ1LltI
The part where the general terms of the AP are defined. I don’t understand how the formula for sum of the natural numbers was arrived at.
Thanks.
In: Mathematics
I suppose you know what APs are and how they work. In this case, we have an AP which are simply the natural numbers: “1,2,3,…”. We want to know two things from this progression:
1) What is the general term of the sequence? That is, what is the n-th number of the sequence?
2) What is the sum of the first n numbers in the sequence? That would be 1+2+3+…+n.
We can answer both questions for any arithmetic progression because there are know formulas for that.
The first one is easy, even if you don’t know the formula. Nonetheless, let’s do it with the formula: The general term “a_n” is
a_n = a_1 + (n-1)*d,
Where a_1 is the first term of the progression (in this case, it’s 1) and d is the diferrence of the progression (each consecutive term differs from the one before by one, because we are adding 1 each time, so d = 1). If we substitute the values in the formula we get
a_n = 1 + (n-1)*1 = 1+n-1 = n <=> a_n = n.
For the second one, there formula is the following: if we call the sum of the first n terms “S_n”, then
S_n = (n/2)*(2*a_1+(n-1)*d).
Again, simply substitute the values and you’ll get the formula for the sum of the natural numbers:
S_n = (n/2)*(2+n-1) = (n/2)*(n+1) (that is what appears in the video).
If you don’t care about using the formulas without knowing why they work, then you’re good to go. If you actually want to know how these formula for the sum was found, then, have a look at this: https://www.youtube.com/watch?v=NEWy3Ckv1Yw
For clarity (I didn’t watch the whole video), are you meaning the formula Sn=(n/2)x[2a1 + (n-1)d]?
If so, it goes like this.
For an arithmetic series, there is an initial seed number a1 (which I’m just going to call a for now) and a difference term d. Each term is made by adding d to the previous term. So the nth term is given by an=a+(n-1)d.
We okay with that?
If so, one thing that we can notice is that we can pair up the numbers. Let’s take the sequence of natural numbers 1,2,3,4,5,6,7,8,9,10,…
Suppose we want to add up the first n terms. We could just do that the long way. Alternatively, we could be clever. If I pair up the numbers starting from both ends, I get that 1+10=11, 2+9=11, 3+8=11, 4+7=11, 5+6=11. So, I now have a bunch of pairs that equal 11.
One way to see this is to take the nth term a+(n-1)d and add it to the first term a. We get 2a+(n-1)d. The next pair would be a2=a+d and a(n-1)=a+(n-2)d. Adding them, we get 2a+(n-1)d, again. Now, how many pairs? Well, we had n numbers which we paired, so we’re going to have n/2 pairs.
So we have n/2 pairs, each of which adds to 2a+(n-1)d, so the total is (n/2)x[2a+(n-1)d]. For the example above, S10 would be 5×11=55.
There’s a famous (probably apocryphal) story about the mathematician Gauss, when he was at school. His teacher was getting annoyed because he was too quick to finish the work. So the teacher said “Fine, if you’re so smart, add up the numbers from 1 to 100.” The teacher looked over a minute later to see Gauss doing nothing. They enquired, and Gauss said “I finished. It’s 5050.” Gauss had used this method, and saw that 1+100=101, 2+99=101,… and 50×101=5050.
For future reference, you might find r/math helpful. There’s a recurring post where people can ask questions 🙂
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