Write a program to print the Fibonacci series 0 1 1 2 3 5 8-------20?

Answer 1

To print the Fibonacci series up to 20, you can write a simple program in a programming language like Python. Here is an example code snippet:

a, b = 0, 1
while a <= 20:
    print(a, end=' ')
    a, b = b, a + b

In this code, we initialize two variables a and b to 0 and 1, respectively. We then use a while loop to iterate and print the Fibonacci numbers until a reaches 20. Inside the loop, we update a and b to generate the next Fibonacci number in the sequence.

Answer 2

fibbonacci starts out as 1,1,2,3,5,8,13,21,34....meaning 1 + 1 = 2, 2 +1 = 3, 3+2 = 5 and so on the solution in C++.

#include <iostream>

using namespace std;

int main()

{

short a = 1;

short b = 1;

short c = 2;

do

{

a = c + b;

cout << a << "\n";

b = a + c;

cout << b << "\n";

c = b + a;

cout << c << "\n";

}while (a && b && c != 34);

return 0;

}

in this i used short int because, i was only going to go up to 34 to keep things short an simple.

I then figured out in what order i would have to figure out my variables on paper. I assigned values to my variables to have a starting point then used a do-while to loop the program. I made sure that my varibles were all checked against 34 so it would have a break point and not run on forever.

Answer 3

Every computer language has different syntax, so writing a set program without knowing the language is infeasible. However, the general algorithm for generating a Fibonacci sequence is as follows. Input -> n; temp1 = 0; temp2 = 1; temp3 = 0; if (n = 0), output: temp1; else if (n = 1), output: temp2; else output: temp1; output: temp2; for loop (base: 0, check: n-1, base++) {temp3 = temp2 + temp1; output: temp3; temp2 = temp1; temp1 = temp3;}

Answer 4

// recursive method - very slow for larger numbers

unsigned long fib (unsigned long n) {

if (n 2) return 1;

return fib (n-1) + fib (n-2);

}

int main (int argc, char* argv[]) {

unsigned long n, f;

for (n = 1; n < 50; ++n) {

f = fib(n);

if (f > 100) break;

printf ("%lu %lu\n", n, f);

}

return 0;

}

Answer 5

#include<stdio.h>

void main()

{

int arr[12],n;

int a=0,b=1,c,i,j,flag=0;

arr[0]=a;arr[1]=b,i=2;

clrscr();

printf("\nEnter the values:");

scanf("%d",&n);

do

{

c=a+b;

arr[i++]=c;

a=b;

b=c;

} while(c<n);

printf("\nNon Fibnoacci series is:");

for(i=0;i<n;i++)

{

flag=0;

for(j=0;j<10;j++)

{

if(i==arr[j])

{

flag=0;

break;

}

else

flag=1;

}

if(flag==1)

{

printf("%d ",i);

}

}

getch();

}

Answer 6

function fib() { local x y z n; x=1; for n in `seq 1 $1` ; do z=$((x+y)); echo -e "$n\t:\t$z"; x=$y; y=$z; done; }

Answer 7

a=0

b=1

c=0

For i=0 to whaterveryouwant

print c

c=a+b

a=b

b=c

next i