To find the sum of this series, we can simplify each term first.

The nth term of the series is given by:
[ \frac{2}{2n-1} \times (2n+1) ]

Now, we can substitute n with 1, 2, 3,..., 49, and sum up all the terms.

The sum of the series is:
[ \sum_{n=1}^{49} \left( \frac{2}{2n-1} \times (2n+1) \right) ]

After substituting the values, we get:
[ \frac{2}{1} \times 3 + \frac{2}{3} \times 5 + \frac{2}{5} \times 7 + ... + \frac{2}{99} \times 101 ]

Now, when n = 49,
[ \frac{2}{2(49)-1} \times (2(49)+1) ]
[ = \frac{2}{97} \times 99 ]

Adding all the terms up, we get:
[ 3 + 5 + 7 + ... + 101 = 2 \times \left( \frac{2}{1} + \frac{2}{3} + \frac{2}{5} + ... + \frac{2}{97} \right) ]
[ = 2 \times \sum_{n=1}^{49} \frac{2}{2n-1} ]

This is a harmonic series, and its sum equals:
[ 2 \times \ln{97} ]
[ = 2 \ln{97} ]