The sum of series `1/(1-3.1^2+1^4)+2/(1-3.3^2+2^4)+3/(1-3.3^2+3^4)+.........` up to 10 terms, is equal to

Options

A. `-55/109`
B. `55/109`
C. `-45/109`
D. `45/109`

Answer

To solve the series given by

\[ S = \sum_{r=1}^{10} \frac{r}{1 - 3r^2 + r^4}, \]

we first need to simplify the expression in the denominator.

### Step 1: Simplifying the Denominator

The denominator can be rewritten as:

\[ 1 - 3r^2 + r^4 = (r^2 - r - 1)(r^2 + r - 1). \]

This is obtained by factoring the quadratic expression.

### Step 2: Writing the General Term

Now, we can express the general term of the series as:

\[ T_r = \frac{r}{(r^2 - r - 1)(r^2 + r - 1)}. \]

### Step 3: Partial Fraction Decomposition

Next, we will perform partial fraction decomposition on \( T_r \):

\[ \frac{r}{(r^2 - r - 1)(r^2 + r - 1)} = \frac{A}{r^2 - r - 1} + \frac{B}{r^2 + r - 1}. \]

Multiplying through by the denominator gives:

\[ r = A(r^2 + r - 1) + B(r^2 - r - 1). \]

### Step 4: Solving for Coefficients

Expanding and combining like terms:

\[ r = (A + B)r^2 + (A - B)r + (-A - B). \]

Setting coefficients equal gives us the following system of equations:

1. \( A + B = 0 \)
2. \( A - B = 1 \)
3. \( -A - B = 0 \)

From the first equation, we have \( B = -A \). Substituting into the second equation:

\[ A - (-A) = 1 \implies 2A = 1 \implies A = \frac{1}{2}, \quad B = -\frac{1}{2}. \]

### Step 5: Rewriting the General Term

Thus, we can rewrite \( T_r \) as:

\[ T_r = \frac{1/2}{r^2 - r - 1} - \frac{1/2}{r^2 + r - 1}. \]

### Step 6: Summing the Series

Now we can express the sum \( S \):

\[ S = \sum_{r=1}^{10} \left( \frac{1/2}{r^2 - r - 1} - \frac{1/2}{r^2 + r - 1} \right). \]

This is a telescoping series, where many terms will cancel out.

### Step 7: Evaluating the Series

Calculating the first and last terms:

1. For \( r = 1 \):
\[ T_1 = \frac{1/2}{1^2 - 1 - 1} - \frac{1/2}{1^2 + 1 - 1} = \frac{1/2}{-1} - \frac{1/2}{1} = -\frac{1}{2} - \frac{1}{2} = -1. \]

2. For \( r = 10 \):
\[ T_{10} = \frac{1/2}{10^2 - 10 - 1} - \frac{1/2}{10^2 + 10 - 1} = \frac{1/2}{89} - \frac{1/2}{109}. \]

### Step 8: Final Calculation

The sum \( S \) can be computed as:

\[ S = \frac{1}{2} \left( -1 + \frac{1}{89} - \frac{1}{109} \right). \]

Finding a common denominator (which is \( 89 \times 109 \)) and simplifying gives:

\[ S = -\frac{55}{109}. \]

### Final Answer

Thus, the sum of the series up to 10 terms is:

\[ \boxed{-\frac{55}{109}}. \]