If `S=sum_(n=2)^(oo) (3n^2+1)/((n^2-1)^3)` then 9/4Sequals

Options

A. `e-2`
B. `e+2`
C. `2e`
D. none of these

Answer

To solve the problem, we need to evaluate the series \( S = \sum_{n=2}^{\infty} \frac{3n^2 + 1}{(n^2 - 1)^3} \) and find \( \frac{9}{4} S \).

### Step-by-step Solution:

1. **Rewrite the Denominator**:
The denominator \( n^2 - 1 \) can be factored as \( (n-1)(n+1) \). Thus, we can rewrite the series:
\[ S = \sum_{n=2}^{\infty} \frac{3n^2 + 1}{((n-1)(n+1))^3} \]

2. **Use the Identity**:
We can use the identity:
\[ (n+1)^3 - (n-1)^3 = 6n^2 \]
This implies:
\[ 3n^2 + 1 = \frac{1}{2} \left( (n+1)^3 - (n-1)^3 \right) \]

3. **Substitute into the Series**:
Substitute this identity into the series:
\[ S = \sum_{n=2}^{\infty} \frac{1}{2} \cdot \frac{(n+1)^3 - (n-1)^3}{((n-1)(n+1))^3} \]

4. **Simplify the Series**:
This can be rewritten as:
\[ S = \frac{1}{2} \sum_{n=2}^{\infty} \left( \frac{1}{(n-1)^3} - \frac{1}{(n+1)^3} \right) \]

5. **Recognize the Telescoping Series**:
The series is telescoping. When we expand it:
\[ S = \frac{1}{2} \left( \left( \frac{1}{1^3} - \frac{1}{3^3} \right) + \left( \frac{1}{2^3} - \frac{1}{4^3} \right) + \left( \frac{1}{3^3} - \frac{1}{5^3} \right) + \ldots \right) \]
Most terms cancel out, leaving us with:
\[ S = \frac{1}{2} \left( 1 + \frac{1}{8} \right) = \frac{1}{2} \cdot \frac{9}{8} = \frac{9}{16} \]

6. **Calculate \( \frac{9}{4} S \)**:
Now, we multiply \( S \) by \( \frac{9}{4} \):
\[ \frac{9}{4} S = \frac{9}{4} \cdot \frac{9}{16} = \frac{81}{64} \]

### Final Result:
Thus, the value of \( \frac{9}{4} S \) is \( \frac{81}{64} \).

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