If n is odd, we’ll have an even number of items (n+1) in each row.īut of course, we don’t want the total area (the number of x’s and o’s), we just want the number of x’s. Notice that this time, we don’t care about n being odd or even – the total area formula works out just fine. We have n rows (we didn’t change the number of rows in the pyramid), and our collection is (n + 1) units wide, since 1 “o” is paired up with all the “x”s. Now for the explanation: How many beans do we have total? Well, that’s just the area of the rectangle. Just like the pairing, one side is increasing, and the other is decreasing. The next row of the pyramid has 1 less x (4 total) and 1 more o (2 total) to fill the gap. Take a look at the bottom row of the regular pyramid, with 5′x (and 1 o).
Let’s say we mirror our pyramid (I’ll use “o” for the mirrored beans), and then topple it over: x o x o o o o oĬool, huh? In case you’re wondering whether it “really” lines up, it does. Well, the sum is clearly 1 + 2 + 3 + 4 + 5. How do we count the number of beans in our pyramid? Sure, we could go to 10 or 100 beans, but with 5 you get the idea. We want to add 1 bean to 2 beans to 3 beans… all the way up to 5 beans. Instead of writing out numbers, pretend we have beans. Different explanations work better for different people, and I tend to like this one better. I recently stumbled upon another explanation, a fresh approach to the old pairing explanation. It works for an odd or even number of items the same! Technique 3: Make a Rectangle Now this is cool (as cool as rows of numbers can be). So we divide the formula above by 2 and get: Notice that we have 10 pairs, and each pair adds up to 10+1.īut we only want the sum of one row, not both. Instead of looping the numbers around, let’s write them in two rows: 1 2 3 4 5 6 7 8 9 10 The above method works, but you handle odd and even numbers differently. It always bugged me that the same formula worked for both odd and even numbers – won’t you get a fraction? Yep, you get the same formula, but for different reasons. And instead of having exactly n items in 2 rows (for n/2 pairs total), we have n + 1 items in 2 rows (for (n + 1)/2 pairs total). Notice that each column has a sum of n (not n+1, like before), since 0 and 9 are grouped. However, our formula will look a bit different. Let’s add the numbers 1 to 9, but instead of starting from 1, let’s count from 0 instead: 0 1 2 3 4īy counting from 0, we get an “extra item” (10 in total) so we can have an even number of rows.
Many explanations will just give the explanation above and leave it at that. What if we are adding up the numbers 1 to 9? We don’t have an even number of items to pair up. Wait - what about an odd number of items?Īh, I’m glad you brought it up. And how many pairs do we have? Well, we have 2 equal rows, we must have n/2 pairs. As the top row increases, the bottom row decreases, so the sum stays the same.īecause 1 is paired with 10 (our n), we can say that each column has (n+1). Instead of writing all the numbers in a single column, let’s wrap the numbers around, like this: 1 2 3 4 5Īn interesting pattern emerges: the sum of each column is 11. Pairing numbers is a common approach to this problem. For these examples we’ll add 1 to 10, and then see how it applies for 1 to 100 (or 1 to any number). Let’s share a few explanations of this result and really understand it intuitively.
#1 2 3 4 5 6 7 8 9 10 next manual#
Manual addition was for suckers, and Gauss found a formula to sidestep the problem: So soon? The teacher suspected a cheat, but no. The so-called educator wanted to keep the kids busy so he could take a nap he asked the class to add the numbers 1 to 100. There’s a popular story that Gauss, mathematician extraordinaire, had a lazy teacher.
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