You have two arithmetic series: 2 + 4 + 6 + ... + 4004 and 1 + 3 + 5 + ... + 4003.
Both of them can be handled in the same way.
t1 is the first term
tn is the nth term (or the last term)
n is the number of terms
d is the common difference between successive terms
S is the sum of the terms.
For the series in the numerator: 2 + 4 + 6 + ... + 4004
t1 = 2
tn = 4004
n is unknown
d = 2
S is unknown
To find n: use this formula: tn = t1 + (n - 1)·d
4004 = 2 + (n - 1)·2
4002 = 2n - 2
4004 = 2n
n = 2002 [there are 2002 terms in this series]
To find S: use this formula: S = n·(t1 + tn) / 2
S = 2002·(2 + 4004) / 2
S = 2002·(4006) / 2
S = 4,010,006
Now, you need to do the same steps for the denominator and find the answer when the numerator is divided by the denominator.