If `1/1^4+1/2^4+1/3^4+...+oo=pi^4/90,` then `1/1^4+1/3^4+1/5^4+...+oo=`

Options

A. `pi^4/96`
B. `pi^4/45`
C. `(89pi^4)/90`
D. `pi^4/90`

Answer

To solve the problem, we need to find the sum of the series \( S = \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \ldots \) given that

\[ \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \ldots = \frac{\pi^4}{90}. \]

### Step-by-Step Solution:

1. **Define the Total Series**:
We know that the sum of the series over all natural numbers is given by:
\[ S_{\text{total}} = \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \ldots = \frac{\pi^4}{90}. \]

2. **Separate the Series into Odd and Even Terms**:
We can separate the total series into odd and even terms:
\[ S_{\text{total}} = S_{\text{odd}} + S_{\text{even}}, \]
where
\[ S_{\text{odd}} = \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \ldots, \]
and
\[ S_{\text{even}} = \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \ldots. \]

3. **Express the Even Series**:
The even series can be factored out:
\[ S_{\text{even}} = \frac{1}{2^4} \left( 1 + \frac{1}{2^4} + \frac{1}{3^4} + \ldots \right) = \frac{1}{16} S_{\text{total}}. \]

4. **Substituting for \( S_{\text{total}} \)**:
Substitute \( S_{\text{total}} \) into the equation:
\[ S_{\text{even}} = \frac{1}{16} \cdot \frac{\pi^4}{90} = \frac{\pi^4}{1440}. \]

5. **Finding \( S_{\text{odd}} \)**:
Now we can find \( S_{\text{odd}} \):
\[ S_{\text{odd}} = S_{\text{total}} - S_{\text{even}} = \frac{\pi^4}{90} - \frac{\pi^4}{1440}. \]

6. **Finding a Common Denominator**:
The common denominator for \( 90 \) and \( 1440 \) is \( 1440 \):
\[ S_{\text{odd}} = \frac{16\pi^4}{1440} - \frac{\pi^4}{1440} = \frac{15\pi^4}{1440}. \]

7. **Simplifying the Result**:
Simplifying \( S_{\text{odd}} \):
\[ S_{\text{odd}} = \frac{15\pi^4}{1440} = \frac{\pi^4}{96}. \]

### Final Answer:
Therefore, the sum \( S = \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \ldots \) is:
\[ \boxed{\frac{\pi^4}{96}}. \]