If `sum_(r = 1)^(oo) (1)/((2r - 1)^2) = (pi^2)/(8)` then the value of `sum_(r = 1)^(oo) (1)/(r^2)` is

Options

A. `(pi^2)/(8) + 1/4 x`
B. `(pi^2)/(4) + 1/3 x`
C. `(pi^2)/(8) - 1/4 x`
D. `(pi^2)/(4) + 2/3 x`

Answer

To solve the problem, we need to find the value of the series \( \sum_{r=1}^{\infty} \frac{1}{r^2} \) given that \( \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} = \frac{\pi^2}{8} \).

### Step 1: Understanding the Given Series
The given series \( \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} \) represents the sum of the reciprocals of the squares of odd integers.

### Step 2: Expressing the Series for All Integers
We know that the series for all integers can be expressed as:
\[ \sum_{r=1}^{\infty} \frac{1}{r^2} = \sum_{r=1}^{\infty} \frac{1}{(2r)^2} + \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} \]
Here, \( \sum_{r=1}^{\infty} \frac{1}{(2r)^2} \) is the sum of the reciprocals of the squares of even integers.

### Step 3: Calculating the Series for Even Integers
The series for even integers can be simplified:
\[ \sum_{r=1}^{\infty} \frac{1}{(2r)^2} = \sum_{r=1}^{\infty} \frac{1}{4r^2} = \frac{1}{4} \sum_{r=1}^{\infty} \frac{1}{r^2} \]

### Step 4: Substituting Back into the Equation
Now we can substitute this back into our equation:
\[ \sum_{r=1}^{\infty} \frac{1}{r^2} = \frac{1}{4} \sum_{r=1}^{\infty} \frac{1}{r^2} + \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} \]
We know from the problem statement that \( \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} = \frac{\pi^2}{8} \), so we can substitute that in:
\[ \sum_{r=1}^{\infty} \frac{1}{r^2} = \frac{1}{4} \sum_{r=1}^{\infty} \frac{1}{r^2} + \frac{\pi^2}{8} \]

### Step 5: Letting \( S = \sum_{r=1}^{\infty} \frac{1}{r^2} \)
Let \( S = \sum_{r=1}^{\infty} \frac{1}{r^2} \). Then we can rewrite the equation as:
\[ S = \frac{1}{4} S + \frac{\pi^2}{8} \]

### Step 6: Solving for \( S \)
To isolate \( S \), we can rearrange the equation:
\[ S - \frac{1}{4} S = \frac{\pi^2}{8} \]
\[ \frac{3}{4} S = \frac{\pi^2}{8} \]
Now, multiply both sides by \( \frac{4}{3} \):
\[ S = \frac{4}{3} \cdot \frac{\pi^2}{8} = \frac{\pi^2}{6} \]

### Final Answer
Thus, the value of \( \sum_{r=1}^{\infty} \frac{1}{r^2} \) is:
\[ \boxed{\frac{\pi^2}{6}} \]