Last week, in Number Sense 034, we developed a formula for figuring out the sum of a series of numbers. We tried out our formula on a couple of examples, and it worked. We also came up with a diagram showing how the formula found the sum for one of the examples. But we didn't prove the formula was true. Two or three examples of it working does not prove it will always work. In order to do that, we must be a bit more rigorous.
That being said, I am not going to do what most mathematicians would recognize as a formal proof. Formal proofs are written in arcane mathematical/logical language, and similar to a programming language or any other language, that language takes a while to learn. However, mathematicians also recognize something called an informal proof, which is written mainly in natural language. We must be careful, though, not to allow any ambiguity to creep into our informal proof, which is always a possibility with something as slippery and prone to misinterpretation as ordinary, day-to-day language.
Here is one of our examples from last week:
2 + 3 + 4 + 5 = 14
It begins with 2, and there are four numbers in the series. Here is a diagram:
We notice that if we take the first row, and add it to the last row, we get 7. We also notice if we take the second row, and add it to the third row, we also get 7.
This gives us half as many rows, and each row is the sum of the first and last number in the series.
When we translate this into a formula, we use a letter to stand for the numbers we use in the example.
Our beginning number is “a”, the amount of numbers in the series is “n” This abstraction, using an unknown number for our actual beginning number, and for the length of the series, is the foundation of our proof. Because the unknown numbers “a” and “n” can be replaced with any number, if we can derive our formula with unknown numbers, we should be convinced that it works for any number we choose.
So, using unknowns, the sum of our series is
Sum = a + (a+1) + … + (a + n - 2) + (a + n – 1)
the dot dot dot in the middle stands in for all the other numbers in the series that I won't bother to write.
The last number in the series adds n -1 (instead of n) because we started the series with “a” all by itself. If we went up to (a + n) we would have one too many numbers in the series. Also notice that, since I don't know how many numbers are unwritten (that dot dot dot part) I begin the series by adding one to the beginning number (counting up) but I end the series by subtracting one from the last number (counting down.)
This is an important step. In order to be convinced of the proof, you must be convinced that a series of n steps that begins at a ends at a + n – 1.
We will write a slightly longer series out:
here is a series (a + 0) + (a + 1) + (a + 2) + (a + 3) + …
we count the number of steps 1 2 3 4 ...
The amount we add to “a” at each step is always one less than the number of steps (I added zero to the first step to show it works even at the beginning number)
We get the next number in the series by adding one to the last number, and by adding one to the number of steps. So the step after step 4 would be step 4 + 1 and it's corresponding number is (a + 3) + 1 or,
step 5, (a + 4)
Since the amount added to “a” in this new step (4) is one less than the number of the step (5) we see we can go one step further and the pattern for the last number remains the same: one less than the step number. Well, if we can go from 4 to 5, we can just as easily go from 5 to 6, 6 to 7, and so on, until we finally reach “n”. And the pattern remains the same with each step, so the pattern will still be true for the last step: step n's number is (a + n – 1).
Now we will add the first number in the series to the last number
a + (a + n – 1) = 2a + n – 1
Then we add the second number to the second to last
(a + 1) + (a + n – 2) = 2a + n – 1
And we get the same result: 2a + n – 1
We keep going like this, and at each step we will always get 2a + n – 1, because each step adds 1 to the first number and subtracts 1 from the second number, or adds zero to the total of the previous two numbers.
Now, when we get to the middle of the series, we have one of two possible situations.
First, the number n might be even, in which case all the numbers neatly pair up, and we simply continue on to the end, adding each number to it's counterpart, until we finally wind up adding the last number to the first number.
Or, the number n might be odd, in which case there is a middle number, which we add to itself, then carry on to the end.
In both cases, we have n sums, each equal to 2a + n – 1, and, in both cases, we have added each number in twice. So the sum of the series, added in only once, would be half what we would get if we simply multiply n by (2a + n – 1)
So, we have
This proves the formula works for any a and any n that we choose.
Have fun in the comments.