Let's analyze the question in the image.

The given problem is asking for the value of the series:

$\sum_{n=0}^{+\infty} \frac{1}{(2n+1)^2}$

We are also provided with the information:

$\sum_{n=1}^{+\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$

Step-by-step breakdown:

The series in question sums over the odd numbers squared, starting from $n = 0$. To connect this with the given sum of squares of all positive integers, we can split the original series of squares of all integers into the sum of even and odd terms.

$\sum_{n=1}^{+\infty} \frac{1}{n^2} = \sum_{n \text{ odd}} \frac{1}{n^2} + \sum_{n \text{ even}} \frac{1}{n^2}$

1. Sum of the even terms:

For even $n$, we substitute $n = 2k$:

$\sum_{n \text{ even}} \frac{1}{n^2} = \sum_{k=1}^{+\infty} \frac{1}{(2k)^2} = \frac{1}{4} \sum_{k=1}^{+\infty} \frac{1}{k^2} = \frac{1}{4} \cdot \frac{\pi^2}{6} = \frac{\pi^2}{24}$

2. Sum of the odd terms:

The remaining part is the sum over odd $n$:

$\sum_{n \text{ odd}} \frac{1}{n^2} = \sum_{n=0}^{+\infty} \frac{1}{(2n+1)^2}$

Since the total sum of squares of all integers is $\frac{\pi^2}{6}$, and the sum of the even terms is $\frac{\pi^2}{24}$, the sum of the odd terms must be:

$\sum_{n=0}^{+\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{4\pi^2}{24} = \frac{\pi^2}{8}$

Thus, the correct answer is D. $\frac{\pi^2}{8}$.

Let me know if you want further details or have any questions.

Here are five related questions to expand on this topic:

1. How is the sum of squares of even integers calculated?
2. What is the value of $\sum_{n=1}^{+\infty} \frac{1}{n^4}$?
3. Can this approach be generalized to other series involving powers of odd and even numbers?
4. What are other famous series that sum to $\pi^2$, like the Basel problem?
5. How do Fourier series connect to summations like this?

Tip: When solving series involving powers of integers, splitting them into even and odd parts can often simplify the process.