#include<iostream>
int main()
{
int x=0, y=1;
std::cout<<x<<" ";
std::cout<<y<<" ";
while( y<1000000 )
{
std::cout<<(y+=x)<<" ";
x=y-x;
}
std::cout<<std::endl;
return(0);
}
Yes, this can be done. For example for Fibonacci series. You will find plenty of examples if you google for the types of series you need to be generated.
i dn't know. haha
#include #include void main() { clrscr() int a=0,b=1,c,i,n; coutn cout
That's the beginning of the Fibonacci series.
It's Fibonacci's
#include<iostream> unsigned fib (unsigned term, unsigned a=0, unsigned b=1) { if (term<1) return a; return fib (--term, a+b, a); } int main() { std::cout << "Fibonacci (1000th term): " << fib (1000) << std::endl; }
(xn+2-1)/(x2-1)
10 print "That is not a question." 20 print "That is a command." 30 end
4 plus 10 plus 16 plus 70 equals 100. To find the sum of this series, simply add all the numbers together.
Find the Sum to n terms of the series 5 5+55+555+ +n Terms
No.
The roots are -1/2 of [ 1 plus or minus sqrt(5) ] . When rounded: 0.61803 and -1.61803. Their absolute values are the limits of the Fibonacci series, or the so-called 'Golden Ratio'.