Key Concepts and Formulas

• Algebraic Manipulation: Techniques like completing the square and difference of squares factorization are crucial for simplifying expressions.
• Partial Fraction Decomposition: Expressing a rational function as a sum or difference of simpler rational functions.
• Telescoping Series: A series where most of the terms cancel out, leaving a sum of a few initial and final terms. The general form is $\sum_{r=1}^{n} (f(r) - f(r+1))$ or $\sum_{r=1}^{n} (f(r+1) - f(r))$.

Step-by-Step Solution

Problem Statement: Find the sum of the series $\frac{1}{1-3 \cdot 1^2+1^4}+\frac{2}{1-3 \cdot 2^2+2^4}+\frac{3}{1-3 \cdot 3^2+3^4}+\ldots$ up to 10 terms.

Step 1: Identify the General Term ($T_r$) The $r$-th term of the series is given by: $T_r = \frac{r}{1 - 3r^2 + r^4}$ Our goal is to simplify this general term to facilitate summation.

Step 2: Manipulate the Denominator We will rewrite the denominator to make it factorizable. The denominator is $1 - 3r^2 + r^4$. We can rearrange this as $r^4 - 3r^2 + 1$. To apply factorization techniques, we can complete the square by adding and subtracting $2r^2$: $r^4 - 3r^2 + 1 = (r^4 - 2r^2 + 1) - r^2$ This step is done to group terms that form a perfect square.

Step 3: Factorize the Denominator using Difference of Squares The expression $(r^4 - 2r^2 + 1)$ is a perfect square, $(r^2 - 1)^2$. So, the denominator becomes: $(r^2 - 1)^2 - r^2$ This is in the form $a^2 - b^2$, where $a = (r^2 - 1)$ and $b = r$. Using the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$: $(r^2 - 1)^2 - r^2 = ((r^2 - 1) - r)((r^2 - 1) + r) = (r^2 - r - 1)(r^2 + r - 1)$ Thus, the general term becomes: $T_r = \frac{r}{(r^2 - r - 1)(r^2 + r - 1)}$

Step 4: Perform Partial Fraction Decomposition (Modified) We want to express $T_r$ as a difference of two terms to create a telescoping series. Let's examine the difference between the two factors in the denominator: $(r^2 + r - 1) - (r^2 - r - 1) = r^2 + r - 1 - r^2 + r + 1 = 2r$ Notice that this difference is exactly twice the numerator. This is a key observation for partial fraction decomposition. We can rewrite $T_r$ as: $T_r = \frac{1}{2} \cdot \frac{2r}{(r^2 - r - 1)(r^2 + r - 1)}$ Now, substitute $2r$ with the difference of the denominator factors: $T_r = \frac{1}{2} \cdot \frac{(r^2 + r - 1) - (r^2 - r - 1)}{(r^2 - r - 1)(r^2 + r - 1)}$ Splitting this into two fractions: $T_r = \frac{1}{2} \left( \frac{r^2 + r - 1}{(r^2 - r - 1)(r^2 + r - 1)} - \frac{r^2 - r - 1}{(r^2 - r - 1)(r^2 + r - 1)} \right)$ $T_r = \frac{1}{2} \left( \frac{1}{r^2 - r - 1} - \frac{1}{r^2 + r - 1} \right)$ This is the desired form for a telescoping series.

Step 5: Calculate the Sum of the First 10 Terms ($S_{10}$) We need to find $S_{10} = \sum_{r=1}^{10} T_r$. $S_{10} = \sum_{r=1}^{10} \frac{1}{2} \left( \frac{1}{r^2 - r - 1} - \frac{1}{r^2 + r - 1} \right)$ $S_{10} = \frac{1}{2} \sum_{r=1}^{10} \left( \frac{1}{r^2 - r - 1} - \frac{1}{r^2 + r - 1} \right)$ Let's write out the terms of the sum to observe the telescoping nature. Let $f(r) = \frac{1}{r^2 - r - 1}$. Then the term inside the summation is $f(r) - f(r+1)$ if we carefully check the second part of the expression. Let's rewrite the second term: $\frac{1}{r^2 + r - 1}$. If we substitute $r+1$ into $r^2 - r - 1$, we get $(r+1)^2 - (r+1) - 1 = (r^2 + 2r + 1) - r - 1 - 1 = r^2 + r - 1$. So, the general term is of the form $\frac{1}{2} (f(r) - f(r+1))$, where $f(r) = \frac{1}{r^2 - r - 1}$.

$S_{10} = \frac{1}{2} \left[ (f(1) - f(2)) + (f(2) - f(3)) + (f(3) - f(4)) + \ldots + (f(10) - f(11)) \right]$ This is a telescoping sum. Most terms cancel out: $S_{10} = \frac{1}{2} [f(1) - f(11)]$ Now, we calculate $f(1)$ and $f(11)$: For $r=1$: $f(1) = \frac{1}{1^2 - 1 - 1} = \frac{1}{1 - 1 - 1} = \frac{1}{-1} = -1$. For $r=11$: $f(11) = \frac{1}{11^2 - 11 - 1} = \frac{1}{121 - 11 - 1} = \frac{1}{109}$.

Substituting these values back into the sum: $S_{10} = \frac{1}{2} [-1 - \frac{1}{109}]$ $S_{10} = \frac{1}{2} \left[ -\frac{109}{109} - \frac{1}{109} \right]$ $S_{10} = \frac{1}{2} \left[ -\frac{110}{109} \right]$ $S_{10} = -\frac{55}{109}$

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful when manipulating the denominator and performing the difference of squares factorization. A small error here will propagate.
• Identifying the Telescoping Pattern: Ensure that the terms you are subtracting are indeed consecutive terms of a function. Writing out the first few terms explicitly is a good way to verify this.
• Handling the Constant Factor: Don't forget the $\frac{1}{2}$ factor that arises from the partial fraction decomposition.
• Calculation of $f(1)$: The first term $f(1)$ involves a negative denominator, which can sometimes lead to sign errors if not handled carefully.

Summary

The problem requires summing a series where the general term can be simplified through algebraic manipulation. By factoring the denominator of the general term $T_r$ using the difference of squares and then applying a modified partial fraction decomposition, we express $T_r$ as a difference of two functions, $f(r) - f(r+1)$. This allows the sum to be evaluated using the telescoping series technique, where intermediate terms cancel out, leaving only the first and last terms of the sum.

The final answer is \boxed{-\frac{55}{109}}. which corresponds to option (B).