Pairing from both ends: a closed form for 1 + 2 + ⋯ + n

0:00Add up the counting numbers from one to ten. You can just do it. One and two is three, and three is six, and four is ten, and so on down the line. Nine additions later you have fifty-five. Now add up the counting numbers from one to a thousand. The same method still works. It just costs nine hundred and ninety-nine additions, and one slip anywhere ruins the lot.

0:28So the question is not whether it can be done. It is whether the answer has a shape. And it does. The shape is easiest to see if you stop reading the list from left to right, and read it from both ends at once. Take the first number and the last number, and add them. One and ten is eleven. Now step inward. Two and nine. That is eleven again.

0:55Three and eight. Four and seven. Five and six. Eleven, eleven, eleven. Ten numbers make five pairs, and every pair comes to the same total. So the whole sum is five elevens, which is fifty-five. The same answer, in one multiplication instead of nine additions. That is not a coincidence. Step one place inward from the left, and the number you are holding goes up by one. Step one place inward from the right, and the number you are holding goes down by one.

1:33Up by one, down by one. The two changes are equal and opposite, so the total cannot move. Every pair inherits the total of the first pair. The pairs agree because they are forced to. And notice what is doing the work. It is the constant step of one. Take a list whose steps are uneven, say one, two, three, ten. Step inward and the left gains one while the right loses seven.

2:03Nothing cancels. The pairs disagree, and the trick has nothing to stand on. There is a snag, and the honest thing is to walk straight into it. Add the counting numbers from one to seven. One and seven is eight. Two and six is eight. Three and five is eight. Three pairs of eight. And four, sitting in the middle, with nobody to pair with. Twenty-four, plus the four left over, is twenty-eight. Correct, but clumsy.

2:35And this is not a rare accident. Of the first hundred runs, exactly fifty hold a term back. Every odd one of them does. Half the time the neat picture has a loose end. A method that works cleanly half the time and needs an apology the other half is not finished. Here is the repair. Write the sum out. Then underneath it, write the same sum backwards. One to ten along the top. Ten down to one along the bottom.

3:09Now every number has a partner directly below it. Nothing is stranded, whatever the length. Read the columns. One and ten, eleven. Two and nine, eleven. Three and eight, eleven. All the way along, ten columns, every one of them eleven. Ten elevens is a hundred and ten. But that is two copies of the sum, not one. You wrote it twice, so you have twice the answer. Twice the sum is a hundred and ten, so the sum is fifty-five.

3:44Now do that with a length of n instead of ten. The top row runs one to n. The bottom row runs n back down to one. Each column holds n plus one, and there are n columns. So twice the sum is n times n plus one. And only now do you halve. The sum is n times n plus one, over two. The doubled form is the one the picture actually gives you.

4:11The halving is a step you take at the end, once the awkward case has already been dealt with. The same argument again, in dots this time. Build a staircase. Six dots in the top row, five in the next, four in the next, down to a single dot. That staircase holds twenty-one dots, which is the sum of one to six. Now take a second staircase, exactly the same, turn it half a turn, and slide it in against the first.

4:44The steps interlock. The jagged edges fit together and the gaps vanish. What is left is a plain rectangle, six across and seven down. Six sevens is forty-two, and forty-two is twice twenty-one. The picture says the same thing as the two rows of writing. Two copies of the sum make a rectangle, and the rectangle is n by n plus one. One loose thread. If twice the sum is n times n plus one, is that always an even number?

5:20It has to be, or the halving would leave a remainder and the whole formula would be suspect. Look at the two numbers being multiplied. They are neighbours. Neighbours never share a parity. One of any two consecutive numbers is even, always. For ten and eleven, it is the ten. For seven and eight, it is the eight. So one of the two factors is even, the product is even, and the halving is exact every single time.

5:51Which is also why halving first is a habit worth breaking. For seven, halving first gives three and a half times eight, and you are carrying a fraction through a problem that has none. Multiply first. Then halve. Every number you write stays whole. Now a run that does not begin at one. Add every number from twenty-five to fifty-eight. The formula only knows about runs that start at one, so use two of them and subtract.

6:24One to fifty-eight comes to seventeen hundred and eleven. Take away the part you do not want, which is one up to twenty-four. Not twenty-five. Twenty-four. Subtract the total up to twenty-five and you have thrown away the twenty-five as well, and your answer is short by exactly that. Three hundred, taken from seventeen hundred and eleven, is fourteen hundred and eleven. And how many numbers is that? Thirty-four, not thirty-three. Fifty-eight minus twenty-four, the same fencepost as the subtraction.

7:05There is a second way to say all of this. Take the average of the two ends. Multiply by how many terms there are. For one to ten, the ends are one and ten, so the average is five and a half. Five and a half, ten times, is fifty-five. For twenty-five to fifty-eight, the average of the ends is forty-one and a half. Forty-one and a half, thirty-four times, is fourteen hundred and eleven. The same answer as before.

7:36It is not a different rule. It is the pairing rule with the halving already folded in. Each pair is worth twice the average, and there are half as many pairs as terms. Does the average of the ends work on any list? Try nought, one, five, six, seven. The ends are nought and seven, the average is three and a half, and there are five terms. Three and a half, five times, is seventeen and a half.

8:06The true total is nineteen. The estimate is short by one and a half. Notice that this list starts with a step of one and ends with a step of one. The ends look perfectly ordinary. It is the middle that misbehaves, and the ends cannot see the middle. The average of the ends is the average of the whole list only when the steps are all the same. Then, and not otherwise.

8:34And one more limit worth stating plainly. Everything here is about the counting numbers, one, two, three, and up. It is not a formula for adding any evenly spaced list you like. Three quick uses. First. Add every two-digit multiple of three. Those are twelve, fifteen, and so on to ninety-nine. Divide each by three and you get the counting numbers four to thirty-three. So the total is three times five hundred and sixty-one, less three times six.

9:12That is sixteen hundred and sixty-five, and there were thirty numbers. Second. Start adding one, two, three, and keep going. When does the running total first pass a thousand? At forty-four you have nine hundred and ninety. At forty-five you have a thousand and thirty-five. So the answer is forty-five. Third, and the prettiest. Which runs of consecutive counting numbers add to exactly one hundred? Two of them. Eighteen through twenty-two, and nine through sixteen. Search every starting point and there are no others.

9:51All of it from one idea. Fold the list onto itself, and the untidy sum turns into a rectangle.