Key Concepts and Formulas

• Algebraic Manipulation: Factoring quadratic expressions and using the difference of squares formula ($a^2 - b^2 = (a-b)(a+b)$) are crucial.
• Telescoping Series: A series where most terms cancel out when summed. The general term $T_n$ is often expressible in the form $f(n) - f(n+1)$ or $f(n-1) - f(n)$.
• Summation Notation: Understanding and manipulating summations ($\sum$) is necessary.

Step-by-Step Solution

1. Identify the General Term ($T_n$): The $n^{th}$ term of the given series is: $T_n = \frac{n}{1 + n^2 + n^4}$

2. Factor the Denominator: The denominator can be rewritten and factored using algebraic identities. $1 + n^2 + n^4 = n^4 + 2n^2 + 1 - n^2 = (n^2 + 1)^2 - n^2$ Applying the difference of squares formula ($a^2 - b^2 = (a-b)(a+b)$) with $a = n^2 + 1$ and $b = n$: $(n^2 + 1)^2 - n^2 = ((n^2 + 1) - n)((n^2 + 1) + n) = (n^2 - n + 1)(n^2 + n + 1)$ So, the general term becomes: $T_n = \frac{n}{(n^2 - n + 1)(n^2 + n + 1)}$

3. Express $T_n$ as a Difference of Two Terms: Our goal is to express $T_n$ in the form $\frac{1}{2} [f(n) - f(n+1)]$ or similar, which is characteristic of a telescoping series. Observe the difference between the two factors in the denominator: $(n^2 + n + 1) - (n^2 - n + 1) = n^2 + n + 1 - n^2 + n - 1 = 2n$ This suggests we can rewrite the numerator $n$ using this difference. We multiply and divide by 2: $T_n = \frac{1}{2} \cdot \frac{2n}{(n^2 - n + 1)(n^2 + n + 1)}$ Substitute $2n = (n^2 + n + 1) - (n^2 - n + 1)$: $T_n = \frac{1}{2} \cdot \frac{(n^2 + n + 1) - (n^2 - n + 1)}{(n^2 - n + 1)(n^2 + n + 1)}$ Now, split the fraction into two terms: $T_n = \frac{1}{2} \left[ \frac{n^2 + n + 1}{(n^2 - n + 1)(n^2 + n + 1)} - \frac{n^2 - n + 1}{(n^2 - n + 1)(n^2 + n + 1)} \right]$ $T_n = \frac{1}{2} \left[ \frac{1}{n^2 - n + 1} - \frac{1}{n^2 + n + 1} \right]$

4. Write the Sum of the First 10 Terms ($S_{10}$): We need to find the sum of the first 10 terms, $S_{10} = \sum_{n=1}^{10} T_n$. $S_{10} = \sum_{n=1}^{10} \frac{1}{2} \left[ \frac{1}{n^2 - n + 1} - \frac{1}{n^2 + n + 1} \right]$ $S_{10} = \frac{1}{2} \sum_{n=1}^{10} \left[ \frac{1}{n^2 - n + 1} - \frac{1}{n^2 + n + 1} \right]$

5. Expand the Sum and Identify the Telescoping Pattern: Let's write out the terms of the sum: For $n=1$: $\frac{1}{1^2 - 1 + 1} - \frac{1}{1^2 + 1 + 1} = \frac{1}{1} - \frac{1}{3}$ For $n=2$: $\frac{1}{2^2 - 2 + 1} - \frac{1}{2^2 + 2 + 1} = \frac{1}{3} - \frac{1}{7}$ For $n=3$: $\frac{1}{3^2 - 3 + 1} - \frac{1}{3^2 + 3 + 1} = \frac{1}{7} - \frac{1}{13}$ ... For $n=10$: $\frac{1}{10^2 - 10 + 1} - \frac{1}{10^2 + 10 + 1} = \frac{1}{91} - \frac{1}{111}$

   The sum is: $S_{10} = \frac{1}{2} \left[ \left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{7}\right) + \left(\frac{1}{7} - \frac{1}{13}\right) + \dots + \left(\frac{1}{91} - \frac{1}{111}\right) \right]$ Notice that the second part of each term cancels with the first part of the next term (e.g., $-\frac{1}{3}$ cancels with $+\frac{1}{3}$). This is the characteristic of a telescoping series.

6. Calculate the Final Sum: After cancellation, only the first part of the first term and the second part of the last term remain. $S_{10} = \frac{1}{2} \left[ 1 - \frac{1}{111} \right]$ To simplify: $S_{10} = \frac{1}{2} \left[ \frac{111 - 1}{111} \right] = \frac{1}{2} \left[ \frac{110}{111} \right]$ $S_{10} = \frac{55}{111}$

Common Mistakes & Tips

• Incorrect Factoring: Ensure the denominator is factored correctly using the difference of squares identity.
• Missing the Factor of 1/2: Forgetting to multiply by 1/2 after expressing the numerator as a difference will lead to an incorrect result.
• Error in Identifying Terms to Cancel: Carefully track which terms cancel. A common mistake is to cancel terms incorrectly, especially at the boundaries of the summation.

Summary

The problem requires finding the sum of the first 10 terms of a given series. The key to solving this problem is to recognize that the general term of the series can be expressed as a difference of two simpler fractions, leading to a telescoping sum. By factoring the denominator and using algebraic manipulation, we rewrote the general term $T_n$ as $\frac{1}{2} \left[ \frac{1}{n^2 - n + 1} - \frac{1}{n^2 + n + 1} \right]$. When summing the first 10 terms, most of these terms cancel out, leaving us with the difference between the first part of the first term and the second part of the tenth term, which simplifies to $\frac{55}{111}$.

Final Answer

The final answer is $\boxed{\frac{55}{111}}$ which corresponds to option (C).