I believe you are familiar with this formula. The German mathematician Gauss is said to have studied the formula at the age of three. To this day, there are many explanations. The most widely used method, which Gauss used in those days, was to add the first and last terms, the second and the penultimate.
For the arithmetic progression , there are
For the arithmetic progression , if , there are
Find . Find
From the above example, we can get
. But this explanation is flawed. When the number of terms is an odd number, there is always a number in the middle that cannot be paired with other Numbers. If I had to figure out the middle number, it would have been too much trouble. Like this example. We can do it the way we did it above, and there are many ways to do it. But let's think about it another way. We can eliminate this trouble by constructing two sequences
, and adding the
term of the first sequence to the
term of the second sequence.
The following picture also proves our derivation. This image comes from Basic Mathematics. In the picture, . We can see that the equation of line in the figure is . It also shows that there is a close relationship between arithmetic progression and the Linear function .
When the teacher talks about this, someone always asks this question. In fact, this problem can be solved by the general formula of arithmetic progression. For any arithmetic sequence ,there are
We combined Linear Function with arithmetic sequence, and set
. So we can also understand
. This needs to do a lot of tedious promotion, here I will not write.
The sum of arithmetic sequence
The above example is an arithmetic sequence with a tolerance equal to one. For arithmetic progression
We can use the above idea to solve the problem. First, we tried to use algebraic methods.
If we substitute , we can get:
There is another way to do this. Thanks to my friend Alex Dixon here, I have modified his method slightly, as follows:
And once again, we can prove it with the pattern above. But it needs to be changed a little bit. I'm sorry that I can't submit pictures here. Those who are interested can draw a picture by yourselves. And you'll see that it becomes a complete trapezoid. And using mathematical induction, of course. Like this:
When , , . There's , so it works.
If , there's .
When , we can get