What is the next number of the sequence 0,1,3,6,10,

Well, when I see a sequence like that, I want to first look at how the difference changes, and make that its own sequence. In this case, the difference goes as the following:

1, 2, 3, 4, 5, 6, 7…

Note: From now on, sequence will refer to the one I just wrote, and sum will refer to the sequence in your question.

This is the classic sequence of natural numbers, so you’re looking at the sum of the first n natural numbers. Now, there are two ways to go about solving this:

First, you can use the good ol’ equations. For the sum of an arithmetic sequence (where the value changes by a constant difference, in this case 1), the sum will be S=(A1+An)∗n/2 , where n is the number of terms in the sequence, A1 is the first number in the sequence, and An is the nth term of the sequence. Now, why does this work? Well, we understand that each term changes by a certain constant, right? It can get hard to sum up every single term individually, but since the change is constant, finding the average value of a set of terms is easy; you only need to take the average of the first and last. Hopefully it makes sense as to why. Now, this would turn your sequence into something along the lines of 3, 3, 3, 3… Where every term is the same. From this, finding the sum is easy! Since you know you have n terms which are all the same, you only need to multiply the average by n! So thus, you have the average multiplied by the number of terms, which one can clearly see becomes (A1+An)∗n/2.

You can quickly plug in numbers for this equation, and thus S=n(n+1)/2. Thus, we’ve found the equation.

On the topic of the equation, you can write it another way, too. Since the sequence is arithmetic, one knows that the difference will always be constant, so at the nth term, the value will be the first term plus (n-1) times the change, since you don’t include the change in the first term. Thus, you could replace An with A1+d(n-1).

Also, it’s important to note that the equation only works with sums beginning at the first term. If you want to sum from the 43rd term to the 65th term, you’d average the 43rd and 65th terms, and multiply by the number of terms contained in the interval you’d like to sum. This would be 65–43+1=23 terms.

Second, you can analyze the sum itself. First of all, after seeing that the change in the terms has a constant increase, it should be immediate to recognize that the sum will be parabolic. I won’t go into detail here, but essentially a parabola is the only function that has constant acceleration. Now, we can start with a basic parabola, and then scale/rotate/transform it until we get the desired function. First: When increasing x by one in a parent function for the parabola (y=x^2), and then one more time, the increase in y from the second change in x will be exactly 2 greater than the first change. We can observe this by looking at the first few points on y=x^2: 0, 1, 4, 9, 16. Find the change between each: 1–0=0, 4–1=3, 9–4=5, 16–9=7, so the new sequence is 1, 3, 5, 7. It’s easy to see that the change here is by 2, so every increase in x by 1 will have the change in y increasing by 2. Now, we can check to see if our sequence is like this: 1, 3, 6, 10, 15: 3–1=2, 6–3=3, 10–6=4, 15–10=5. This is a change by 1, which is only half of the change in two we should be seeing. To fix this, we can halve the slope, since halving the rate of change also halves the change in slope, too. This turns our equation into y=x^2/2: 0, 1/2, 2, 9/2, 8: 1/2–0=1/2, 2–1/2=3/2, 9/2–2=5/2, 8–9/2=7/2. So now, we have fixed the rate of change of the slope, but the slope itself is shifted off by 1/2 to our slope (adding n/2 to the actual graph). Thus, we can add n/2 to our equation, getting S=(n^2+n)/2, which is the same equation we found in the first method.

I don’t feel that my explanation on the second method was as strong, so feel free to comment about it for any questions/suggestions!