1, golden section
With the increase of the number of items in the series, the ratio of the former item to the latter item approaches the golden section value of 0.6 18039887. ...
2. Rectangular
Fibonacci sequence is related to the generation of rectangular region, from which a property of Fibonacci sequence can be derived. The sum of squares of the first few terms of Fibonacci sequence can be regarded as squares of different sizes, which can be combined into a large rectangle because of Fibonacci recurrence formula. So the sum of the areas of all the small squares is equal to the area of the big rectangle. Then the following identities can be obtained:
3, mantissa cycle
One digit of Fibonacci sequence: 60-step cycle
1 1235,83 145,94370,774 15,6 1785.38 190,
99875,27965, 16730,33695,49325,729 10…
Furthermore, the last two digits of Fibonacci sequence are cycles of 300 steps, the last three digits are cycles of 1500 steps, the last four digits are cycles of 15000 steps, and the last five digits are cycles of 150000 steps.
4. Fibonacci series in film and television works.
Fibonacci sequence is widely known in Europe and America, so it often appears in the popular art of movies. For example, it appears as an important symbol and plot clue in the popular Da Vinci Code, and it is also a random question asked at the job fair for shopkeepers in Magic Toy City. It can be seen that this series is as popular as the golden section.
Fibonacci series also appeared in TV series, for example, in the fifth time of the Japanese TV series Kao Shen, Yisi did the last math problem in the national mock exam, which was quoted countless times in the popular American TV series mystery files by Fox, and even served as one of the design elements of the whole drama poster.
5. Yang Hui Triangle
Align Yang Hui's triangle to the left to form an arrangement as shown in the figure, and add the numbers on the same diagonal to get 1, 1, 2, 3, 5, 8, ...
The formula is as follows:
f⑴=C(0,0)= 1 .
f⑵=C( 1,0)= 1 .
f⑶=C(2,0)+C( 1, 1)= 1+ 1=2 .
f⑷=C(3,0)+C(2, 1)= 1+2=3 .
f⑸=C(4,0)+C(3, 1)+C(2,2)= 1+3+ 1=5 .
f⑹=C(5,0)+C(4, 1)+C(3,2)= 1+4+3=8 .
f⑺=c(6,0)+c(5, 1)+c(4,2)+c(3,3)= 1+5+6+ 1= 13。
……
f(n)=C(n- 1,0)+C(n-2, 1)+…+C(n- 1-m,m)(m & lt; =n- 1-m)