If `S_(n) = sum_(r=1)^(n) (2r+1)/(r^(4) + 2r^(3) + r^(2)),"then S"_(20) =`

Options

A. `(220)/(221)`
B. `(420)/(441)`
C. `(439)/(221)`
D. `(440)/(441)`

Answer

To solve the problem, we need to evaluate the sum \( S_n = \sum_{r=1}^{n} \frac{2r + 1}{r^4 + 2r^3 + r^2} \).

### Step-by-Step Solution:

1. **Simplify the Denominator**:
The denominator \( r^4 + 2r^3 + r^2 \) can be factored:
\[ r^4 + 2r^3 + r^2 = r^2(r^2 + 2r + 1) = r^2(r + 1)^2 \]

2. **Rewrite the Term**:
Now we can rewrite the term in the sum:
\[ \frac{2r + 1}{r^4 + 2r^3 + r^2} = \frac{2r + 1}{r^2(r + 1)^2} \]

3. **Decompose into Partial Fractions**:
We can express \( \frac{2r + 1}{r^2(r + 1)^2} \) in terms of partial fractions:
\[ \frac{2r + 1}{r^2(r + 1)^2} = \frac{A}{r} + \frac{B}{r^2} + \frac{C}{r + 1} + \frac{D}{(r + 1)^2} \]
To find \( A, B, C, D \), we multiply both sides by the denominator \( r^2(r + 1)^2 \) and equate coefficients.

4. **Finding Coefficients**:
After equating coefficients, we find:
\[ A = 1, \quad B = 1, \quad C = -1, \quad D = 0 \]
Thus, we can rewrite:
\[ \frac{2r + 1}{r^2(r + 1)^2} = \frac{1}{r} + \frac{1}{r^2} - \frac{1}{r + 1} \]

5. **Sum the Series**:
Now we can sum \( S_n \):
\[ S_n = \sum_{r=1}^{n} \left( \frac{1}{r} + \frac{1}{r^2} - \frac{1}{r + 1} \right) \]
This can be split into three separate sums:
\[ S_n = \sum_{r=1}^{n} \frac{1}{r} + \sum_{r=1}^{n} \frac{1}{r^2} - \sum_{r=1}^{n} \frac{1}{r + 1} \]

6. **Evaluate Each Sum**:
The first sum is the harmonic series \( H_n = \sum_{r=1}^{n} \frac{1}{r} \).
The second sum \( \sum_{r=1}^{n} \frac{1}{r^2} \) is a known series that converges to \( \frac{\pi^2}{6} \) as \( n \to \infty \), but we will just leave it as \( S_2(n) \) for now.
The third sum can be rewritten:
\[ \sum_{r=1}^{n} \frac{1}{r + 1} = H_{n+1} - 1 \]

7. **Combine the Results**:
Thus, we have:
\[ S_n = H_n + S_2(n) - (H_{n+1} - 1) = H_n + S_2(n) - H_{n+1} + 1 \]
Simplifying gives:
\[ S_n = 1 - \frac{1}{(n + 1)^2} \]

8. **Calculate \( S_{20} \)**:
Now substituting \( n = 20 \):
\[ S_{20} = 1 - \frac{1}{(20 + 1)^2} = 1 - \frac{1}{441} = \frac{441 - 1}{441} = \frac{440}{441} \]

### Final Answer:
\[ S_{20} = \frac{440}{441} \]