Answer / s.vyasaraj

#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 / chue kimcheang

C++
void main()
{
int i,n,sum=0;
cout<<"N=";
cin>>n;
for(i=1;i<=n;i=i+2)
sum=sum+i;
cout<<"\n1+3+5+7+....+"<<n<<"="<<sum;
}

C program....
void main()
{
int i,n,sum=0;
printf("N=");
scanf("%d",&n);
for(i=1;i<=n;i=i+2)
sum=sum+i;
printf("\n1+3+5+7+....+%d=%d",n,sum);
}

Answer / mazharul haque

#include<stdio.h>
#include<conio.h>
void main()
{
int n,i,s=0;
printf("ENTER A RANGE ");
scanf("%d",n);
for(i=1; i<=n; i=i+2)
{
s=s+i;
printf("%d+",i);
}
printf("=%d",s);
getch();
}

Answer / mazharul haque

#include<stdio.h>
#include<conio.h>
void main()
{
int n,i,s=0;
printf("ENTER A RANGE ");
scanf("%d",& n);
printf("1");
for(i=3; i<=n; i=i+2)
{
s=s+i;
printf("+%d",i);
}
printf("=%d",s+1);
getch();
}

Answer / naresh kumar

In the answer Printf function is worng here the value for %
d is not mentioned it would like

printf("the series is =%d", sum);

Answer / mariaalex007

#include"stdio.h"
#include"conio.h"
void main()
{
int s=0;
for(int i=1;i<100;i++)
{
if(i%2==0)
continue;
s=s+i;
}
printf("The answer is...%d",s);
getch();
}

Answer / @pravin.08

main()
{
int i=1,n;
printf("enter n ");
scanf("%d",&n);
while(i<=n)
{
printf("%d+",i);
i+=2;
}
}

Answer / nagarajan

Simplest method to do it ..

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

Answer / gaurav pitaliya

#include<stdio.h>
#include<conio.h>
void main()
{
int i,s=0;
clrscr();
for(i=0; i<=10; i++)
{
s=s+(2*i)+1;
}
printf("The Sum of series is= %d",s);
getch();
}

Answer / aoyn

#include<iostream.h>
#include<conio.h>

void main()
{
clrscr();
int i,n,sum=0;
cout<<“1+2+3+……+n”;
cout<<“nEnter the value of n:”;
cin>>n;

for(i=1;i<=n;++i)
sum+=i;
cout<<“nSum=”<<sum;
getch();
}

char* f() return "hello:"; void main() {char *str=f(); }

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