Gaussian summation

Gauss, a famous German mathematician, was brilliant in his childhood. When he was at school, one day the teacher asked the students to calculate: 1+2+3+4+…+99+ 100.

After the teacher finished the problem, the whole class was immersed in calculation, but Gao Xiao Jones quickly figured out that the answer was equal to 5050. It turns out that little Gauss found through careful observation that:

1+ 100=2+99=3+98=…=49+52=50+5 1

1 ~ 100 can be divisible by 50 logarithms, and the sum of each logarithms is equal. So, little Gauss cleverly calculated this problem as:

( 1+ 100)× 100÷2=5050。

Extended data:

Gauss's story:

Gauss is the son of an ordinary couple. His mother is the daughter of a poor stonemason. Clever as she is, she has no education and is almost illiterate. Before becoming Gauss's father's second wife, she was a maid. His father used to be a gardener, a foreman, an assistant to a businessman and an appraiser of a small insurance company. It has become an anecdote that Gauss was able to correct his father's debt account when he was three years old. He once said that he could do complicated calculations in his head.

When I was a child, Gauss's family was poor, and his father found it useless to study, but Gauss still liked reading. It is said that when he was a child, his father would tell him to go to bed after dinner in winter to save fuel, but when he went to bed, he would hollow out the inside of the radish and put it into a cotton roll, and then continue reading when the lamp was on.

When Gauss 12 years old, he began to doubt the basic proof in element geometry. When he was 16 years old, it was predicted that a completely different geometry would be produced outside Euclidean geometry, that is, non-Euclidean geometry. He derived the general form of binomial theorem, successfully applied it to infinite series and developed the theory of mathematical analysis.

arithmetical series formula

Arithmetic progression formula an=a 1+(n- 1)d

The first n terms and formulas are: Sn=na 1+n(n- 1)d/2.

If the tolerance d =1:sn = (a1+an) n/2.

If m+n=p+q, then: am+an=ap+aq exists.

If m+n=2p, then: am+an=2ap.

All the above n are positive integers. And Sn, prime minister a 1, final term an, tolerance d, term number n.