How to write a program to print 1 2 3... series in C?

Answer 1

main(){

int temp,prev,i;

for(i=2,prev=1;prev<=256;i++){

printf("%d ",prev);

temp=i;

i=i*prev;

prev=temp;

}

}

Answer 2

hi here is the program for u //program to print the series 1,-4,7,-10,13,-16 etc...

#include<stdio.h>

#include<math.h>

void main()

{

int i,n,sign=1;

printf("Enter the number of terms to which u want to print the series\n");

scanf("%d",&n);

for(i=1;i<=(1+(n-1)*3);i=i+3)

{

sign=pow((-1),(i+1));

printf("%d\t",(sign*i));

}

} here the sign is changing for every even term so we can use (-1)^(i+1), so when i=even then power of -1 will be odd...hence it will be -1.now in the for loop part in the middle section i have put an A.P. formula for number of terms, in arithmatic progression tn=a+(n-1)*d where tn=nth term, a=first term d= the difference of the series if u still have any confusions plz feel free to buzz me...thru mail adishredder@Yahoo.co.in

Answer 3

...

char ch;

int number = 0;

do

{

cout << number << " ";

number++;

} while (ch != s);

//When you press 's' the loop will quit.

...

Answer 4

// Here are three ways to answer your question. Hope it's not a homework assignment! // -----Method 1------ // #include int main() { printf("1121231234"); return 0; } // -----Method 2------ // #include int main() { int i, j; for(i=1; i

Answer 5

#include <stdio.h>

int main (void) { puts ("1 12 123 1234 12345"); return 0; }