Answer / shiva

for(int a=0;a<10;a++){
(a&1)?printf("%d\n",a):printf("\n");
}

Answer / nasif

#include<stdio.h>
int main()
{
int n,i,d=1,sum=0;
scanf("%d", &n);
for(i=1;i<=n;d=d+2){
sum=sum+d;
i++;
}
printf("summation is %d", sum);
return 0;
}

Answer / meheka

#inclined <stdio.h>
int main ()
{
int I=1,n,sum=0;
scanf("%d",&n);
while(i<=n)
sum=sum+I;
{
printf("%d",sum);
I=I+2;
}
return 0:

Answer / beetha poulose

#include<include>
int main ()
{
int j,i,n,sum=0;
Print("enter n");
scanf ("%d",&n);
For (i=1,j=1; j<=n; i=i+5,j++)
{
Sum=sum +i;
}
Prinrf ("sym=%d",sum);
Return 0;
}

Answer / nishant

#include<stdio.h>
#include<conio.h>
void main()
{
int n,i=1,sum=0;
clrscr();
printf("enter the value for n:");
scanf("%d",&n);
while(i<=n)
{
sum=sum+i;
i=i+2;
}
printf("the series is =%d");
getch()
}

Answer / aditya parashar

#include<iostream.h>
void main ()
{int s,n;
cout <<"For finding sum of series
1+3+5.. till n terms enter value of n: ";
cin>>n;
s=n*n;
cout<<"
sum of series 1+3+5+.. till n
terms is: "<<s;
}
[Note:- only output matters]

Answer / medo

//*I know this is a different program but i want you to try
it!*//



#include <stdio.h>
void main()
{
int i,j;

for(i=1;i<=4;i++)
{
for(j=1;j<=8;j++)
{printf("* ");}
printf("\n");

for(j=1;j<=8;j++)
{printf(" *"); }
printf("\n");
}
}

Answer / parthibanselvaraj

the series seems to be infinite . so you need to restrict
it to some ceiling.

who was the present cheif governor of reserve bank of india

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