VB Perbonacci sequence, Perbonacci sequence

Do you want to ask how to use a vb program to represent this series of numbers?

Each term of the sequence is equal to the sum of the first two terms.

an=a(n-1)+a(n-2) n>2

private sub command1_click()a = 1

b = 1sum = 2

for i = 3 to 20

t = a + b

a = bb = t

sum = sum + b

next i

Item 20 is:" & b & "and for" & sumend sub

。。。an=a(n-1)+a(n-2) n>2

Each term of the sequence is equal to the sum of the first two terms.

3 is getting closer and closer (5-1) 2

This can be proved by the definition of a sequence.

f(n+1) = f(n) +f(n-1) let f(n+1) f(n) = k

Then f(n) f(n-1) = k

Then k 2 = k + 1

Solving this equation yields k = ( 5+1) 2

So the ratio of adjacent terms is (5-1) 2

The recursive formula is a(n+2)=a(n+1)+a(n), so the eigenequation is x 2 = x + 1

Solve x1=(1+root number 5) 2, x2=(1- 5) 2 So the general formula is an=c[(1+root number5) 2] n-d[(1-root number5) 2] n, where c and d are the coefficients to be determined.

Substituting the known terms into it determines c and d.

Give it a try

If we let the general term of the desired sequence be a(n), then since the difference between the adjacent terms of the series is the Perbonacci sequence, we can get the younger push: a(n+1)-a(n)=f(n) From this second push formula, we can get the following formula: a(2)-a(1)=f(1).

a(3)-a(2)=f(2)

a(4)-a(3)=f(3)

a(n-1)-a(n-1)=f(n-1)

a(n)-a(n-1)=f(n-1)

By adding the above formula to the left and right, we can easily get:

a(n)-a(1)=f(1)+f(2)+.f(n-1)=s(n-1) (is the sum of the first n-1 terms of the Fibonacci sequence), then at this point, our problem is transformed into the problem of finding the sum of the first n terms of the Fibonacci sequence, and the process of the sum of the first n terms of the Fibonacci sequence will be given below

We already know that for the Fibonacci sequence f(n) we have such a recursive formula, namely: f(n+1)=f(n)+f(n-1)(), from which we get: f(n-1)=f(n+1)-f(n)s, from which we get:

f(1)=f(3)-f(2)

f(2)=f(4)-f(3)

f(3)=f(5)-f(4)

f(n-1)=f(n+1)-f(n)

f(n)=f(n+2)-f(n+1)

Add the above n formulas left and right to get:

f(1)+f(2)+f(3)+.f(n)=f(n+2)-f(2)=f(n+2(-1=s(n).This equation shows that the first n terms of the Fibonacci sequence are exactly the n+2 terms of the Fibonacci sequence minus 1

Now, we have solved the summation problem of the Fibonacci sequence, and from the previous equation, we know that a(n)-a(1)=s(n-1), since a(1)=0So: a(n)-0=a(n)=s(n-1)=f(n+1)-1= 5-1

The Fibonacci sequence refers to such a sequence of numbers 、...This sequence starts with the third term, each of which is equal to the sum of the first two terms.

The formula for the Fibonacci sequence is shown in the figure.

Starting with item 3, each item is the sum of the previous 2 items.

Analysis: Fabonacci sequence.

Recursive relation: f(n-2) + f(n-1) = f(n) general term formula: