To find the ratio of the sum of odd terms and the sum of even terms in an arithmetic progression (A.P.) with (2n + 1) terms, we need to understand the structure of the A.P. and how the terms are distributed.

Let's assume the first term of the A.P. is 'a' and the common difference is 'd'. The (2n + 1) terms will be:

a, a + d, a + 2d, a + 3d, ..., a + nd, a + (n + 1)d

We can observe that the odd terms will be the terms with an odd index (1st, 3rd, 5th, etc.), and the even terms will be the terms with an even index (2nd, 4th, 6th, etc.).

The sum of the odd terms can be calculated by adding all the odd-indexed terms:

Sum of odd terms = a + (a + 2d) + (a + 4d) + ... + [a + (2n)d]

Similarly, the sum of the even terms can be calculated by adding all the even-indexed terms:

Sum of even terms = (a + d) + (a + 3d) + (a + 5d) + ... + [a + (2n + 1)d]

Let's simplify these sums to find the ratio.

Sum of odd terms:
Sum of odd terms = a + (a + 2d) + (a + 4d) + ... + [a + (2n)d]
= (2n + 1)a + (2 + 4 + ... + 2n)d
= (2n + 1)a + 2(1 + 2 + ... + n)d
= (2n + 1)a + 2n(n + 1)d

Sum of even terms:
Sum of even terms = (a + d) + (a + 3d) + (a + 5d) + ... + [a + (2n + 1)d]
= na + (1 + 3 + 5 + ... + (2n + 1))d
= na + n(n + 1)d

Now, let's calculate the ratio of the sum of odd terms to the sum of even terms:

Ratio = (Sum of odd terms) / (Sum of even terms)
= [(2n + 1)a + 2n(n + 1)d] / [na + n(n + 1)d]
= (2n + 1)a / na + (2n + 1)d / (n + 1)d
= (2n + 1)/n + (2n + 1)/(n + 1)

Simplifying this ratio further, we get:

Ratio = (2n + 1)(n + 1) / n(n + 1)
= (2n + 1)n + (2n + 1) / n(n + 1)
= 2n + 1 / n

Therefore, the ratio of the sum of odd terms to the sum of even terms is n : (n - 1), which matches option B.