If `1/(1^2)+1/(2^2)+1/(3^2)+ tooo=(pi^2)/6,t h e n1/(1^2)+1/(3^2)+1/(5^2)+` equals `pi^2//8` b. `pi^2//12` c. `pi^2//3` d. `pi^2//2`

Options

A. `pi^2//8`
B. `pi^2//8`
C. `pi//3`
D. `pi^2//2`

Explanation

To solve the problem, we need to find the value of the series:

\[ S = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots \]

We know from the problem statement that:

\[ \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \ldots = \frac{\pi^2}{6} \]

### Step 1: Write the series for all integers
The series for all integers can be expressed as:

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

### Step 2: Separate the series into odd and even terms
We can separate this series into two parts: the series of odd integers and the series of even integers.

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

### Step 3: Express the even terms
The series for even integers can be expressed as:

\[ \sum_{k=1}^{\infty} \frac{1}{(2k)^2} = \sum_{k=1}^{\infty} \frac{1}{4k^2} = \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{1}{4} \cdot \frac{\pi^2}{6} = \frac{\pi^2}{24} \]

### Step 4: Substitute back into the equation
Now we can substitute this back into our equation for the sum of all integers:

\[ \frac{\pi^2}{6} = \sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} + \frac{\pi^2}{24} \]

### Step 5: Solve for the sum of odd integers
Let \( S = \sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} \). Then we have:

\[ \frac{\pi^2}{6} = S + \frac{\pi^2}{24} \]

### Step 6: Isolate \( S \)
To isolate \( S \), we can subtract \( \frac{\pi^2}{24} \) from both sides:

\[ S = \frac{\pi^2}{6} - \frac{\pi^2}{24} \]

### Step 7: Find a common denominator
To perform the subtraction, we need a common denominator. The common denominator of 6 and 24 is 24. Thus, we rewrite \( \frac{\pi^2}{6} \):

\[ \frac{\pi^2}{6} = \frac{4\pi^2}{24} \]

Now substituting back:

\[ S = \frac{4\pi^2}{24} - \frac{\pi^2}{24} = \frac{3\pi^2}{24} \]

### Step 8: Simplify
Simplifying \( S \):

\[ S = \frac{3\pi^2}{24} = \frac{\pi^2}{8} \]

### Conclusion
Thus, the value of the series \( \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots \) is:

\[ \frac{\pi^2}{8} \]

### Final Answer
The correct option is **(a)** \( \frac{\pi^2}{8} \).

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