Let
1 1 1
P = --- + ---- + .....+ ------
1*2 3*4 99*100
1 1 1
Q = ------ + ------ + ....+ ------
51*100 52*99 100*51
Express P/Q in its lowest terms, that is, gcd(P,Q)=1
*** Adapted from a problem appearing at Argentina Mathematical Olympiad in 2014
Generalize into P(n) = 1/[1*2] + 1/[3*4] + ... + 1/[(4n-1)*(4n)]
and Q(n) = 1/[(2n+1)*(4n)] + 1/[(2n+2)*(4n-1)] + ... + 1/[(4n)*(2n+1)]
Then this problem will be the case when n=25.
Break down P(n) by fraction decomposition:
P(n) = 1 - 1/2 + 1/3 - 1/4 + ... + 1/(4n-1) - 1/(4n)
Now expand each negative term -1/(2x) into 1/(2x)-1/x:
P(n) = 1 + [1/2 - 1] + 1/3 + [1/4 - 1/2] + ... + 1/(4n-1) + [1/(4n) - 1/(2n)]
Now rearrange terms by denominators:
P(n) = [1 - 1 + 1/2 - 1/2 + ... + 1/(2n) - 1/(2n)] + [1/(2n+1) + 1/(2n+2) + ... + 1/(4n)]
The first sum is zero, then rearrange the second sum in pairs:
P(n) = [1/(2n+1) + 1/(4n)] + [1/(2n+2) + 1/(4n-1)] + ... + [1/(3n) + 1/(3n+1)]
Then add the pairs:
P(n) = (6n+1)/[(2n+1)*(4n)] + (6n+1)/[(2n+2)*(4n-1)] + ... + (6n+1)/[(3n)*(3n+1)]
Double both sides, on the left multiply by 2 and on the right copy the sum in reverse order:
2*P(n) = {(6n+1)/[(2n+1)*(4n)] + (6n+1)/[(2n+2)*(4n-1)] + ... + (6n+1)/[(3n)*(3n+1)]} + {(6n+1)/[(3n+1)*(3n)] + ... + (6n+1)/[(4n-1)*(2n+2)] + (6n+1)/[(4n)*(2n+1)]}
Factor 6n+1 on the right:
2*P(n) = (6n+1)*{1/[(2n+1)*(4n)] + 1/[(2n+2)*(4n-1)] + ... + 1/[(4n-1)*(2n+2)] + 1/[(4n)*(2n+1)]}
Now the second term of the product on the right is Q(n). Then after some simple rearrangement
P(n)/Q(n) = (6n+1)/2
So then to answer the specific problem at hand: P/Q = (6*25+1)/2 = 151/2.
Solution
