If ` a = sum_(r=1)^(oo) 1/(r^(2)), b = sum_(r=1)^(oo) 1/((2r-1)^(2)),` then `(3a)/b` is equal to _____

Answer

To solve the problem, we need to find the value of \( \frac{3a}{b} \), where:

- \( a = \sum_{r=1}^{\infty} \frac{1}{r^2} \)
- \( b = \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} \)

### Step 1: Calculate \( a \)

The series \( a \) is a well-known series, which converges to:

\[ a = \sum_{r=1}^{\infty} \frac{1}{r^2} = \frac{\pi^2}{6} \]

### Step 2: Calculate \( b \)

The series \( b \) can be expressed as:

\[ b = \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} = \sum_{n=1}^{\infty} \frac{1}{n^2} - \sum_{r=1}^{\infty} \frac{1}{(2r)^2} \]

The first term is \( a \), and the second term can be simplified as follows:

\[ \sum_{r=1}^{\infty} \frac{1}{(2r)^2} = \sum_{r=1}^{\infty} \frac{1}{4r^2} = \frac{1}{4} \sum_{r=1}^{\infty} \frac{1}{r^2} = \frac{1}{4} a \]

Thus, we can write:

\[ b = a - \frac{1}{4} a = \frac{3}{4} a \]

### Step 3: Calculate \( \frac{3a}{b} \)

Now substituting \( b \) into the expression \( \frac{3a}{b} \):

\[ \frac{3a}{b} = \frac{3a}{\frac{3}{4}a} = \frac{3a \cdot 4}{3a} = 4 \]

### Final Answer

Thus, the value of \( \frac{3a}{b} \) is:

\[ \boxed{4} \]