program to get 0.9,0.99,0.999,0.9999.....nth terms.

//program to get 0.9,0.99,0.999,0.9999.....nth terms.
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
float n,j, k=0.9;
clrscr();
printf("enter number of terms\n");
scanf("%f",&n);
for(j=1;j<=n;j++)
{
printf("%f,",k);
k=k+0.9/(pow(10,j));
}
getch();
}
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logics in mind for this:-
->0.9,0.99,0.999............. can be written as
             0.9,0.9+0.09,0.99+0.009.......
->         for second term 0.99, it can be written as 0.9+0.09
 ->further it can be as(taking only two terms)
            0.9+( 0.9+0.9/10)...
->it can be as
          0.9+(0.9+0.9/10^{1)...
->now we have to convert it in a formula as}
        k=k+0.9/(pow(10,j)).here k=0.9 is taken
->and to display all the terms we have used loop

or

we can apply following technique:
//getting 0.9,0.99,0.999.....nth term
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
float k=9,n,p=10,j,dis,m=1;
clrscr();
printf("enter number of terms\n");
scanf("%f",&n);
for(j=1;j<=n;j++)
{
dis=k/(pow(10,j));
printf("%f,",dis);
m++;
k=pow(p,m)-1;
}
getch();
}

->take 9 as a fixed value
->we enter no. of terms for 'n'.
->to display 0.9, we divide it (9) by 10<sup>1
->to get 0.99, we take here 99, and for this we use 100-1 concept. we do this using k=pow(p,m)-1
        here pow finds power of 10(p).
->in next execution we get 0.99 and so on..

-> this is how loop continues and we get all terms

we can use different technique for same program.</sup>

         similarly we can also get 
1)9,99,999,9999......
2)2,22,222,2222....
3)5,55,555,5555......
4)0.99999,0.9999,0.999,0.99,0.9
    for this,
      there are 5 nines so it can be written as
         (100000-1)/10⁵,(10000-1)/10⁴........
         (10⁵-1)/10⁵,(10⁴-1)/10⁴........  
       now make a formula to get this one. And it is easy task...