 Pascal considered the problem of summing the powers of terms in an arithmetic sequence. So remember that an arithmetic sequence is one generated by a formula ak equals a0 plus kd, where d is the common difference. His rule appears in the arithmetic triangle. The rule is more complicated to state than to apply, so we'll introduce it by way of examples. Let's begin with a simple one. Let's find the sum of the series 5 plus 8 plus 11 plus 14 plus 17. So Pascal's approach is the following. To begin with, we note that the terms of the series are raised to power 1, and the common difference is 3. Since we're summing the first powers of an arithmetic sequence with common difference 3, Pascal found the difference between the second powers of n plus 3 and n. And that's easy enough to compute. Now, why don't we compute the second powers when we're interested in the sums of the first powers? Pascal made the following observation. If we consider the differences between the squares of successive terms of the sequence, our formula gives us a way to compute these differences. So 20 squared minus 17 squared is 6 times 17 plus 9. 17 squared minus 14 squared is 6 times 14 plus 9. And we can compute the other differences. And at this point, something very interesting happens. If we add everything up on the left-hand side, all but the first and last terms cancel out. And if we collect the terms on the right-hand side, we see we have our sum, 5 plus 8 plus 11 plus 14 plus 17, and a couple other terms. And this gives us an equation that we could solve for the sum that we want. And so we find, of course, you say that's a lot of work to finding this sum because we could have just added 5 plus 8 plus 11 plus 14 plus 17 directly. But this provides a way to sum longer sequences and other powers. For example, suppose we want to find the sum of 5 plus 9 plus all the way up to 85. And so we make a few observations. Since the terms are raised to power 1 and have common difference 4, we find the difference of the squares of two numbers separated by 4. And that will be... And so now we find a whole bunch of differences. So we'll start with 89 squared minus 85 squared. That's, in some sense, the next term that we didn't include. 85 squared minus 81 squared and so on. If we add all the terms in the left column, everything cancels except the first and last. And the right side gives us the sum we want plus a whole bunch of 16s. And the only actually difficult part of this is determining the number of rows that we have. So let's take a look at that. So the numbers here are 5, 9, and so on, up to 85. And if you notice, every row is 4 more than the preceding. That's because our common difference is 4. And this top row, 85, well that's really 5 plus 24s. And so that means there's 20 plus 1, 21 rows. So we have 21 16s. And so this difference of squares has this nice expression. And we can solve it for the sum that we want. More importantly, Pascal's method gives us a way to find the sums of the powers of terms in an arithmetic sequence. So let's find the sum of the series 5 squared plus 8 squared and so on. Since the terms with common difference 3 are raised to the second power, we'll consider the difference between the third powers of n plus 3 and n. And we'll find our differences. So the next term that we didn't include is 20 squared. So we'll start by finding 20 cubed minus 17 cubed. And all the rest, adding the left side, adding the right side. And that gives us... Now our previous result would have given us this sum, it turns out to be 55, and so we can solve for the sum of the squares. Now we might focus on the sums of consecutive integers. So since our common difference of the consecutive integers is 1 and the power is 1, we'll consider the difference n plus 1 squared minus n squared. That's just 2n plus 1. And let's consider those differences k plus 1 squared minus k squared all the way down to 2 squared minus 1 squared. And these will be... And if I add the terms up on the left-hand side, everything but the first and last terms cancel. And on the right-hand side we'll get... And we can rearrange and solve this for our sum of the integers. So doing that gives us... What if we wanted to find the sum of the squares of consecutive integers? So because they're consecutive integers, the common difference is 1. And because we're looking at the sum of the squares, we'll want to look at the difference of the consecutive cubes, n plus 1 cubed minus n cubed, which will be... And adding up our differences gives us k plus 1 cubed minus 1 cubed is equal to this expression. And we just found what the sum of the consecutive integers is. So we can rearrange our terms and using our previous result, obtain a formula for the sums of the successive squares of the integers. Pascal goes on to note that those familiar with the doctrine of indivisibles can apply these summations to finding area. In particular, if we apply the rules to continuous quantities, the sum of the lines to the square on the greatest is 1 to 2. The sum of the squares on the lines to the cube of the greatest is 1 to 3. The sum of the cubes of the lines to the fourth power of the greatest is 1 to 4, and so on. So Pascal is essentially describing the following procedure. Consider the curve y equal x to the n. Now if I draw a bunch of ordnance spaced by some small amount o, the sums of all the ordnance will be a sum of powers of an arithmetic sequence with common difference o. And the sum is 1 over n plus 1 of the n plus first power of the last ordnance. But this only shows a few of the ordnance if we put in all of the ordnance, then, at least by the doctrine of indivisibles, the sum of all of the ordnance is the area under the curve. And so what Pascal is describing is a very non-rigorous description of a Riemann sum and a very, very, very, very rough precursor to our integration formula.