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The ancient algorithm of trigonometric number
How did ancient China and Greeks calculate this sum?

Pythagoras, a Greek mathematician 2,400 years ago, called such numbers 1, 1+2, 1 +2+3, 1+2+3+4 and so on. He and his disciples use 1 sphere to represent 1, and use the following figure to represent the number of triangles:

We use Sn to represent the value of1+2+3+…+n. To know the number of Sn, we can imagine another Sn (represented by a white ball here), turn it upside down and put it together with the original Sn; We get a rhombus (Figure 2, where n equals 4), and there are always n rows, and each row has n+ 1 spheres, so there are n(n+ 1) spheres. These are two SNs, so one SN should be n(n+ 1)÷2.

Coincidentally, China people also use this method to find the value of Sn. Yang Hui, a mathematician in the Song Dynasty, thought it was a sharp pile of straw, with a bundle above it and a bundle from top to bottom. If you know the number of bales at the bottom, you can calculate the total number of all bales. One of the questions he asked was: "There are a bunch of osmanthus flowers today, one above and eight below. Ask * * * how many bundles? A: 36 bundles. " His calculation method is the same as the above explanation.

Pythagoras and his disciples discovered a property of trigonometric numbers: the sum of any two consecutive trigonometric numbers is a square number. Figuratively speaking, it is:

Readers can prove the above properties with formulas.

An easy-to-think question: Can 1 2+2 2+3 2+…+N 2 and 1 3+3 3+…+N 3 also find a simple formula to calculate their sum?

It is said that people who found the law of buoyancy in the bathhouse forgot that they were still streaking in the street and shouted "Eureka! Eureka! " I found it! I found it! ) The Greek scientist Archimedes (Archimedes, 287- 2 BC12) knew for a long time that the formula of these two sums is:

1^2+2^2+3^2+…+n^2=n(n+ 1)(2n+ 1)÷6

1^3+2^3+3^3+…n^3=( 1+2+…+n)^2

However, the Greek mathematicians after Archimedes wanted to know the formula of the sum of 1 4+2 4+3 4+…+n 4, but they could do nothing. This formula was not known until 1000 years later, when the Arab mathematician arshi settled in 10 century.

Let's ask a question: Is there a general formula to express the sum of 1 m+2 m+…+n m for any m≥3?

The French mathematician Fermat solved this problem.

1636, the French mathematician p Fermat wrote a letter to a friend, which said, "I solved the most beautiful problem in arithmetic." The question he said is the one asked above.

Fermat discovered this formula:

He naturally thought, whether?

To his surprise, it turned out to be like this.

He also set out from here and got a beautiful formula:

For P≥2, the following equation is an identity.

This formula can be used to solve the formula that mathematicians have been trying to solve for more than a thousand years. Look at the simplest case first, that is, p=2:

The formula on the left can be expanded and written.

Knowing the formula of ∑r, we can substitute the formula of ∑r2.

Know the formula of ∑r2, and then consider the case of p=3, because

The value of Σ RM. Fermat skillfully solved the problem.