Consider a denominator of r;

It has proper fractions:

1/r, 2/r, ...., (r-1)/r

Their sum is: (1 + 2 + ... + (r-1))/r

The numerator of this sum is

1 + 2 + ... + (r-1)

Which is an Arithmetic Progression (AP) with r-1 terms, and sum:

sum = number_of_term(first + last)/2

= (r-1)(1 + r-1)/2

= (r-1)r/2

So the sum of the proper fractions with a denominator or r is:

sum{r} = ((r-1)r/2)/r = ((r-1)r/2r = (r-1)/2

Now consider the sum of the proper fractions with a denominator r+1:

sum{r+1} = (((r+1)-1)/2

= ((r-1)+1)/2

= (r-1)/2 + 1/2

= sum{r) + 1/2

So the sums of the proper fractions of the denominators forms an AP with a common difference of 1/2

The first denominator possible is r = 2 with sum (2-1)/2 = ½;

The last denominator required is r = 100 with sum (100-1)/2 = 99/2 = 49½;

And there are 100 - 2 + 1 = 99 terms to sum

So the required sum is:

sum = ½ + 1 + 1½ + ... + 49½

= 99(½ + 49½)/2

= 99 × 50/2

= 2475

The sum of the series

a + ar + ar2 + ... is a/(1 - r) for |r| < 1

Sum of r = summer

one fourth the sum of r and ten is identical to r munus 4

void main() { int num,r,sum=0; clrscr(); printf("enter the number\n"); scanf("%d",&num); while(num!=0) { r=num%10; sum=sum+r; num=num/10 } printf("The sum of individual digit of given number is=%d",sum); getch() }