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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

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โˆ™ 2017-09-07 17:10:58
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โˆ™ 2017-09-07 12:44:10

It is 2475.

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Q: What is the sum of all the positive proper fractions with denominators less than or equal to 100?
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