It won't go so far

Short solution:

$\displaystyle \dfrac {(100001 + 199999) \times 50000}{(1 + 99999) \times 50000} = \dfrac {300000 \times 50000}{100000 \times 50000} = 3$

Long solution:

In the numerator,

$\displaystyle (a_k + a_{n - k})_{numerator}$ is a constant value:

$\displaystyle 100001 + 199999 = 100003 + 199997 = (a_k + a_{n - k})_{numerator} = 300000$

And there are

$\displaystyle \dfrac{(a_n - a_1) + 1}{2} = \dfrac {100000}{2} = 50000$

iterations of this constant value, so

$\displaystyle 100001 + 100003\ + \ ...\ + 199997 + 199999 = 300000 \times 50000$

Similarly, in the denominator $\displaystyle (a_k + a_{n - k})_{denominator}$ is also constant.

Notice that all of the terms in the denominator are just the terms in the numerator $\displaystyle + \ 100000$ , so

$\displaystyle (a_k + a_{n - k})_{denominator} = (a_k + a_{n - k})_{numerator} - (100000 \times 2)$ .

$\displaystyle (a_k + a_{n - k})_{denominator} = 100000$

And since the terms in the denominator are all just shifted from those in the numerator by $100000$ , it has the same number of iterations of $\displaystyle a_k + a_{n - k}: 50000$ .

$\displaystyle \dfrac {100001 + 100003 + 100005\ + \ ... \ + 199999}{1 + 3 + 5\ + \ ... \ + 99999} = \dfrac {300000 \times 50000}{100000 \times 50000} = 3$