practising C, C++ and JAVA programming with examples: 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 10¹
^{->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.}

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...

menu

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