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