To solve this problem, we need to find the value of n for which the sum of the series is equal to 100/101.

Let's first rewrite the series in sigma notation:

∑ (2/(2n - 1)(2n + 1)) from n=1 to k

Now, we need to find the value of k for which the sum of the series is equal to 100/101. Let's denote the sum of the series as S:

S = 2/3 + 2/15 + 2/35 + ..... + 2/(2k - 1)(2k + 1)

Now, we can rewrite the series in terms of k:

S = ∑ (2/(2n - 1)(2n + 1)) from n=1 to k

S = 2∑ (1/(2n - 1)(2n + 1)) from n=1 to k

S = 2(1/3 - 1/5 + 1/5 - 1/7 + ... + 1/(2k - 1) - 1/(2k + 1))

S = 2(1/3 - 1/(2k + 1))

Now, we can set this equal to 100/101 and solve for k:

2(1/3 - 1/(2k + 1)) = 100/101

2/3 - 2/(2k + 1) = 100/101

2/(2k + 1) = 2/101

2k + 1 = 101

2k = 100

k = 50

Therefore, the value of n for which the sum of the series is equal to 100/101 is 50.