The calculation power of Fibonacci sequence

1. The general term formula of Fibonacci sequence is
an = √ 5 / 5 [(1 + √ 5) / 2] ^ n - √ 5 / 5 [(1 - √ 5) / 2] ^ n, let BN = √ 5 / 5 [(1 + √ 5) / 2] ^ n, CN = √ 5 / 5 [(1 - √ 5) / 2] ^ n
then an = BN CN, {BN} is an equal ratio sequence with the common ratio of (1 + √ 5) / 2, {CN} is an equal ratio sequence with the common ratio of (1 - √ 5) / 2, The first n items of BN and the first n items of BN and the first n items of BN and the first n items of BN and the first n items and the first n items and the first n items and the first n items of CN and the first n items and the first n items and the first n items of CN and the first n items and the first n items and the first n items and the first n items and the first n items and the first n items and the first n items and the first n items and the first n of the first n (5 / 5 [(1 + [5 [[5) (1 + [5 [5]) (5 [5 [5 [5) (1 + [5) (1 + [5) (1 + [5) (1 + [5) / 5) / 2]] * (1 - [(1 - (1 + [5 [1 + [1 + [5 [1 + [5 [1 + [5 [5 [1 + [1 + [1 + [5) (1 + [5) (1 + [5) / 5) / 5) (1 + [5) / 5) / 5) / 5)] (5] = a1 + A2 +... + an = b1-c1 + b2-c2 +... + BN CN = BN CN
= (3 √ 5 + 5) ([(1 + √ 5) / 2] ^ n-1) / 10 - (3 √ 5-5) ([ (1-√5)/2]^n-1)/10
={(3√5+5)([(1+√5)/2]^n-1)-(3√5-5)([(1-√5)/2]^n-1)}/10

2. This can be summed up by the general term formula
the general formula is an = (1 / √ 5) * {[(1 + √ 5) / 2] ^ n - [(1 - √ 5) / 2] ^ n}
sum Sn for 0-n (A0 = 0, for the convenience of calculation, it has no effect on the result)
use the sum formula of equal ratio sequence. (1-A (1-A (n + 1) (1-A) (1-A (n + 1) (1-A) (1-A) (1-A) (1-A (n + 1) (1-A) (1-A (n + 1) (1-A (n + 1) (1-A) (1-A (n + 1) (1-A (n + 1) (1-A (n + 1) (1-A (n + 1) (1-A) (1-A) (1-A (n + 1 + 1) (1-1) (1-1) (1-1-1-1) (1-1-1-1) (1-1-1-1-1-1-1) / 5 (5-5-5-5-5-5-5-5-5-5-5-5-5-5-1) / 5-1 / 5-1 / 5-1) / 2) ((((1 / 5-5-5-5-5-5-5-5-5-5-5-5-5-5-5-5-5 this is the solution of {5 * {((1 + √ 5) / 2) ^ (n + 2) - ((1 - √ 5) / 2) ^ (n + 2)} - 1
and we found that by the general term formula,
Sn = a (n + 2) - 1, I verified that this formula is correct. A (n + 2) is the N + 2 term of Fibonacci sequence
in fact, we can easily prove the correctness of this formula by mathematical inction. If you don't know, ask me again.

3. From an + 2 = an + 1 + an
with an + 2-An + 1-an = 0
construct the characteristic equation x2-x-1 = 0,
let its two roots be p, Q have PQ = - 1, P + q = 1

let's prove that {an + 1-pan} is an equal ratio sequence with Q as the common ratio

for the convenience of derivation, let A0 = 1, still satisfy an + 2 = an + 1 + an

an + 1-pan
= an + an-1 - Pan
= (1-p) an-pqan-1
= Q (an-pan-1)
so: {an + 1-pan} is an equal ratio sequence with Q as the common ratio< < br / < br /
< br / < br / < br / < br / < br / < br / < br / < br /
< br / < br / < br /
< br / < br /
< br / < br / < br / < br / < br /
< br /
< br / < br / < br / < br /


as a result of P = (1-5) / 2, q = (1-5) / 2, q = (1 + (1-5) / 2, q = (1 + (1 + tick5) / 5) / 2, and Q-P = 5, Q-P = 5, Q-P = 5, so < br / < br /
an = (1 / {(1 / {[1 / {[1 / √ [5] 5) {[5] {[5] {[5] {[[[(1 / {[5] {[[(1 it is suitable for the above general formula.

4. Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21...
if f (n) is the nth term of the sequence (n ∈ n +). Then this sentence can be written in the following form:
F (1) = f (2) = 1, f (n) = f (n-1) + F (n-2) (n ≥ 3)
obviously, this is a linear recursive sequence< The first derivation method of general term formula is as follows:

x ^ 2 = x + 1
the characteristic equation of linear recurrence sequence is: X1 = (1 + √ 5) / 2, X2 = (1 - √ 5) / 2.
then f (n) = C1 * X1 ^ n + C2 * x2 ^ n
∫ f (1) = f (2) = 1
C1 * X1 + C2 * x2
C1 * X1 ^ 2 + C2 * x2 ^ 2
the solution is C1 = 1 / √ 5, C2 = - 1 / √ 5
F (n) = (1 / √ 5) * {[(1 + √ 5) / 2] ^ n - [(1 - √ 5) / 2] ^ n} [√ 5 denotes radical 5]
derivation method 2 of the general term formula: let R, s
be a constant such that f (n) - R * f (n-1) = s * [f (n-1) - R * f (n-2)]
then R + S = 1, - rs = 1
n ≥ 3, There are all < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br /
F (n-n-n-2) - R * f (n-1-1) - R * f (n-1-1) - R * f (n-1 (n-1-1-1) - R * f (n-1 (n-1) - R * f (n-1 (n-2) - R * f (n-2) - R * f (n-2 (n-n-2) - R * f (n-n-1 (n-1) is (n-2)] < br / < br / < br / < br / < br /
F (n (n (n (n-n-n-n 2) - R * f (1)]
∵ s = 1-r, The results of the above formula can be reced to: < br /
the following formula can be reced to: < br /
if you want to get the following: < br /
F (n) = s (n-1) (n-1) + R * f (n-1) (n-1) + R (n-1 (n-1) + R (n-1) + R (n-1) + R (n-1-1) + R * f (n-1) (n-1) < br /
then: < br /
then: < br / < br /
then: < br / < br /
then: < br / < br / < br / < br /

< br /
< the following < br / < br / < br / < br > < br /


< the following < < < < br / < br > < br / < br / < br(n-3) +... + R ^ (n-2) * s + R ^ (n-1) * f (1)
= s ^ (n-1) + R * s ^ (n-2) + R ^ 2 * s ^ (n-3) +... + R ^ (n-2) * s +R ^ (n-1)
(this is the sum of the items of an equal ratio sequence with S ^ (n-1) as the first term, R ^ (n-1) as the end term and R / s as the tolerance)
= [S ^ (n-1) - R ^ (n-1) * r / S] / (1-r / s)
= (s ^ n - R ^ n) / (S-R)
a solution of R + S = 1, - rs = 1 is s = (1 + √ 5) / 2, r = (1 - √ 5) / 2
then f (n) = (1 / √ 5) * {[(1 + √ 5) / 2] ^ n - [(1 - √ 5) / 2] ^ n}

5. Solution: share a solution. The general formula F (n + 2) = AF (n + 1) + BF (n), f (1) = f (2) = 1. The problem is a special case of a = b = 1. It is concluded that f (n) = C1 (x1) ^ n + C2 (x2) ^ n, which is derived from the "inverse application" of the generation process of characteristic equation. The process is to add "- XF (n + 1)" on both sides of F (n + 2) = f (n + 1) + F (n), and let an = f (n + 1) - XF (n), an + 1 = (1-x) an + [(1-x) x + 1] f (n) To construct {an} as an equal ratio sequence, let the coefficient of F (n) (1-x) x + 1 = 0, that is, the characteristic equation x ^ 2 = x + 1]. In this way, the first term of an is f (2) - XF (1) = 1-x, and the common ratio is (1-x). The results show that: an = f (n + 1) - x1f (n) = (1-x1) ^ n (1), an = f (n + 1) - X2F (n) = (1-x2) ^ n (2), and (x2-x1) f (n) = (1-x1) ^ n - (1-x2) ^ n can be obtained from (1 - 2). In this problem, 1-x1 = X2, 1-x2 = x1, x2-x1 = - 1 / √ 5, the expression of F (n) is sorted out. For reference.

6. This is a bit difficult

7.

The answer is yes

In fact, arbitrary:

A (n + 2) = AA (n + 1) + ban form, There are two main methods to solve this kind of problems: 1. Undetermined coefficient method 2. Characteristic equation method

the following figure is the proof of the completeness and characteristic equation of undetermined coefficient method to solve this kind of problems

I will explain an application of characteristic equation method for LZ with a special example

{1, 1, 2, 3, 5, 8, 13, 21, ...}

it is not difficult to find that this sequence has two very significant characteristics: A1 = A2 = 1 and an = a (n-1) + a (n-2)

in fact, this is the famous Fibonacci sequence. From the third term, the following term is the sum of the first two terms

which is equivalent to a (n + 2) = AA (n + 1) + ban form of A, Through the "characteristic equation" method proved in the figure below, it can be seen that:

solving the characteristic equation x ^ 2 = x + 1 of an = a (n-1) + a (n-2), we can get

x1, where x2 is (1 + with 5) / 2 and (1-with 5) / 2 respectively

then there is an= α[ 1 + and 5) / 2] ^ n+ β[ 1-5) / 2] ^ n

where α And β It can be solved by substituting A1 and A2 α= 1 / and 5, β=- 1 / followed by 5

that is an = (1 / followed by 5) {[(1 + followed by 5) / 2] ^ n - [(1-followed by 5) / 2] ^ n}

this is the general formula of Fibonacci sequence

then for a (n + 2) = AA (n + 1) + ban form of three adjacent recursion formula

only need to solve its characteristic equation x ^ 2 = ax + B

① has only one real root: {an / (x ^ n)} is the arithmetic sequence

undeterminable coefficient, let an = [A1 + (n-1) D] x ^ (n-1)

and then determine the value of d by A2

② has two unequal real roots:

undeterminable coefficient let an= α( x1)^n+ β( X2) ^ n

is determined by A1 and A2 α and β If there's anything else that LZ doesn't understand, you can ask

I hope my answer will be helpful to you

8. The latter is the sum of the first two. The denominator of complex fraction is always greater than 1, so the value is always less than 1
and the numerator always takes the previous denominator, except that when the first numerator denominator is 1, the value is equal to 1 / 2, and the later values are greater than 1 / 2
and the denominator of complex fraction denominator is always the same each time the complex fraction is calculated, The numerator is always the sum of the previous numerator and denominator
which fully conforms to the expansion law of Fibonacci sequence

so what is the value of the simplest infinitely continuous fraction
what is the limit of the ratio of two consecutive terms of Fibonacci sequence
suppose: x = 1 / (1 + 1 / (1 + 1) / (1 +...))
obviously: x = 1 / (1 + x)
that is: x ^ 2 + X-1 = 0
x = (√ 5-1) / 2 = 0.618... (rounding off the negative value)
this is the golden ratio, which is also the limit of the ratio of two consecutive terms of Fibonacci series
this is the reason why the landlord said: "it is getting closer to the golden ratio"
the so-called "with the increase of N, the gap between the two numbers becomes smaller and smaller", in fact, it is getting closer and closer to the limit

then why are "any two numbers added continuously" like this
the golden section ratio is actually a problem of the ratio between the middle and the outside:
the so-called ratio between the middle and the outside is to divide a given line segment into two parts, so that one part is the middle of the ratio between the whole line segment and the other part
if the longer segment is set to x, the shorter segment is 1-x
therefore, x ^ 2 = 1 * (1-x) [where "1" represents the whole segment]
that is, x ^ 2 + X-1 = 0, which is completely consistent with the above equation for solving the simplest infinitely continuous fraction
note that the whole segment here is represented by 1, which means that the golden section ratio has nothing to do with the actual length of the segment
similarly, For the expansion of Fibonacci sequence, if we consider the ratio of the front and back terms, then it doesn't matter which two numbers are added from
because it is always the ratio of the large number in the two numbers to the sum of the two numbers, which has the same meaning with the golden section's external ratio
besides, the first ratio is not the sum ratio, All other ratios are always between 0.5 and 1
if the first two numbers are not the same, then: m, N, M + N, M + 2n, 2m + 3N, 3M + 5N,...
it can be seen that they are expanded according to the law of Fibonacci sequence. Of course, this is a general understanding. Strict proof depends on relevant data
think again, if the first two numbers of Fibonacci sequence are 1 and 2? It's different
it's not the same. Except for the first item, it's no different
if the first two numbers are the same, then m, m, 2m, 3M,... Is actually a Fibonacci sequence,
each number is only m times different, which does not affect the ratio of two consecutive terms. Moreover, from the third term, the coefficients before a just constitute the Fibonacci sequence
starting from the second term, the coefficients before B just constitute the Fibonacci sequence
so, The formula of the item that is the following formula that is the formula that is the following formula that is the following formula that is the formula that is the following formula that is the following formula that is the following formula that is the following formula that is the formula that is the following formula:
the coefficient before the number a of the number a is the following formula:
the coefficient before the number a of the number a is the following formula:
the coefficient before the number number a is the coefficient of the number a in the first number of the first number a (the coefficient of the first number of the first number of the first number of the first number of the first number a (n-2) - [(1 + (1 + {(1 + {(1 + (1 + {(1 + {(1 + {(1 + {(1 / / \\\√ 5) * {[(1 + √ 5) / 2] ^ (n)- 1 - [(1-√5)/2]^n-1}*b

9. The number of digits of Fibonacci sequence is 60 cycles, and four digits in each cycle are 2: < br > 2013 ÷ 60 = 33... 33 < br > if the remainder is 33, then only the third digit in this interval is 2, < br > 33 × 4+1
=132+1
=133< Br > A: among the first 2013 items of Fibonacci series, the last digit of 133 items is 2

10. Just take 305 or 320 at Jinsha station in red light