We looked at series. Let's now look at some examples where we can calculate the results without having to add the terms one by one. I mean using a shortcut, using a formula. The first of such examples is the arithmetic series. So let's look at a simple one. So we have 1, 2, 3, 4, 5, and so on. It is an arithmetic progression with common difference one and first term one. The series will be the sum of all of these sums. And say we're going to add them all the way to 100, just because I really like the story of the mathematician Gauss that worked out the formula for this when he was a primary school boy. I'll leave you to find out more. So we want to add all these numbers, one all the way to 100. The clever thing that Gauss did was to note that the first and the last term, 1 and 100, add up to the same amount as two and the term before 100, or 99. They all add up to 101. And the same goes on to all the other pairs of numbers in the middle. So, 1 plus 100 is 101, the other pair 2 plus 99 is 101. And 3 with 98 is 101, and so on. All we need to find out is how many pairs are there. So the last pair to add up will be the middle one, 50 + 51. So there are 50 pairs altogether, which means that the sum above there is calculated with 101 added to itself 50 times. So, 101 times 50, Which is 5,050. Large number. Let's take a rule out of this example. So what we did was we added the first and the last term. We took that sum, in this case it was 101, and we multiply that by the number of pairs. How could we calculate the number of pairs? Well, we've got 100 numbers. So, it's 100 divided by two, it's 50, it's the number of pairs. So we can generalize the pattern here to any arithmetic progression. So we did the first term added with the last term, and then times the number of terms all divided by 2. So in the example we just saw a1 was 1, an was 100, and we have 100 terms. And we divided this by 2, so we had 101 times 50, the 50 came from that, okay? The sum of an arithmetic progression. In the example we saw, we had an even number of numbers to add up, which meant we had a number of pairs. What if we had an odd number of numbers to add in the series? This formula would still work if we're adding an odd number of terms of an arithmetic progression. Let's look at two examples of this formula in action. Want to calculate the sum of odd numbers, From 1 to 100. So we want to add 1 + 3 + 5, and so on, all the way to 99, okay? All the odd numbers. We can use the arithmetic series formula if this is actually an arithmetic series, and it is. It is an arithmetic progression with first term 1, and the common difference 2, therefore the general term will be 1 + (n- 1) 2. And so the sum of the sequence when we run through all the odd numbers is going to be the sum when n is 1 all the way to n equals 50. Let's check, when n is 50, we got 49 times 2 plus 1, and that gets us 99, okay? So this is given by the first term plus the last, times 50 which is the number of terms divide by 2. So that means I do 1 plus 99 which is 100 times 50 divide by 2. That's 50 times 50 which is 2,500. Premeet. If I wanted the sum of even numbers, From 1 to 100, I'm looking at adding 2 + 4 and so on, all the way to 100. I want to write it as a sum of an arithmetic progression. And let's write with bn, and bn is going to be 2 + (n- 1) times the common difference which is still two in this case. And the last term is 100, and I achieve that when n is 50, because I start with two and I add 49 2's to it to get 100, so it's also the sum to 50. And therefore, this result is the first term plus the last, that's 102 x 50 divide by 2, and that turns out to be 51 x 50, which is 2,550.