prakash pankaj

• Dec 1st, 2012

 

Code
#include<stdio.h>


#include<conio.h>


int main()


{


int r=2,i,a;


for(i=1,a=1;a<10,i<10;a++,i++)


{


printf("%d",(a+pow(r,i));


}


return 0;


}

shweta gupta

 Profile Answers by shweta gupta 

• Jun 25th, 2014

 

Code
#include<iostream>


using namespace std;


int main()


{


int temp=3;


for(int i=0;i<10;i++){


        cout<<temp<<endl;


        temp=temp*2-i;


 


 


}


  


    return 0;


}

nira

• Jul 1st, 2014

 

Much easier approach,
this is a logic question, not programming

Code
    


    #include<stdio.h>


    #include<math.h>


    int main()


    {


    int a=1,r=2,i;


    for (i=0;i<=10;i++)


    printf(",%d ", a * pow(r,i) + i);


    return 0;


    }