Key Concepts and Formulas

• Partial Fraction Decomposition: Expressing a rational function as a sum or difference of simpler rational functions.
• Telescoping Series: A series where most terms cancel out when the sum is computed, leading to a simplified form. The general form is $\sum_{r=1}^{n} (a_r - a_{r+1})$ or $\sum_{r=1}^{n} (a_{r-1} - a_r)$.
• Algebraic Manipulation: Recognizing and applying algebraic identities like the difference of squares: $a^2 - b^2 = (a-b)(a+b)$.

Step-by-Step Solution

Step 1: Identify the General Term ($T_r$) We are given a series, and we need to find the sum of its first 20 terms. Let's write down the general term, $T_r$, for the $r$-th term of the series: $T_r = \frac{4r}{4 + 3r^2 + r^4}$

Step 2: Manipulate the Denominator to Facilitate Factoring Our goal is to rewrite the denominator $4 + 3r^2 + r^4$ in a form that allows for easier factorization, ideally leading to a difference of terms that can be used for telescoping. We can rearrange and complete the square: $4 + 3r^2 + r^4 = r^4 + 4r^2 + 4 - r^2$ The first three terms, $r^4 + 4r^2 + 4$, form a perfect square: $(r^2 + 2)^2$. So, the denominator becomes: $(r^2 + 2)^2 - r^2$ This is a difference of squares, where $a = r^2 + 2$ and $b = r$. Applying the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$: $(r^2 + 2 - r)(r^2 + 2 + r) = (r^2 - r + 2)(r^2 + r + 2)$ Thus, the general term can be written as: $T_r = \frac{4r}{(r^2 - r + 2)(r^2 + r + 2)}$

Step 3: Express $T_r$ as a Difference of Two Terms (Partial Fraction Decomposition) To create a telescoping series, we need to express $T_r$ as a difference of two terms. We observe that the numerator $4r$ can be related to the difference between the two factors in the denominator. Let's consider the difference between the quadratic terms: $(r^2 + r + 2) - (r^2 - r + 2) = r^2 + r + 2 - r^2 + r - 2 = 2r$ The numerator is $4r$, which is $2 \times (2r)$. So, we can write $4r = 2 \times [(r^2 + r + 2) - (r^2 - r + 2)]$. Now, substitute this back into the expression for $T_r$: $T_r = \frac{2[(r^2 + r + 2) - (r^2 - r + 2)]}{(r^2 - r + 2)(r^2 + r + 2)}$ We can split this into two fractions: $T_r = 2 \left[ \frac{r^2 + r + 2}{(r^2 - r + 2)(r^2 + r + 2)} - \frac{r^2 - r + 2}{(r^2 - r + 2)(r^2 + r + 2)} \right]$ Simplifying each fraction: $T_r = 2 \left[ \frac{1}{r^2 - r + 2} - \frac{1}{r^2 + r + 2} \right]$

Step 4: Calculate the Sum of the First 20 Terms ($S_{20}$) We need to find the sum of the first 20 terms, $S_{20} = \sum_{r=1}^{20} T_r$. $S_{20} = \sum_{r=1}^{20} 2 \left[ \frac{1}{r^2 - r + 2} - \frac{1}{r^2 + r + 2} \right]$ $S_{20} = 2 \sum_{r=1}^{20} \left[ \frac{1}{r^2 - r + 2} - \frac{1}{r^2 + r + 2} \right]$ Let's examine the structure of the terms inside the summation. Let $f(r) = \frac{1}{r^2 - r + 2}$. Then, $f(r+1) = \frac{1}{(r+1)^2 - (r+1) + 2} = \frac{1}{r^2 + 2r + 1 - r - 1 + 2} = \frac{1}{r^2 + r + 2}$. So, the term inside the summation is $f(r) - f(r+1)$. This confirms it's a telescoping series of the form $\sum_{r=1}^{n} (f(r) - f(r+1))$.

Let's write out the first few terms and the last term to see the cancellation: For $r=1$: $2 \left[ \frac{1}{1^2 - 1 + 2} - \frac{1}{1^2 + 1 + 2} \right] = 2 \left[ \frac{1}{2} - \frac{1}{4} \right]$ For $r=2$: $2 \left[ \frac{1}{2^2 - 2 + 2} - \frac{1}{2^2 + 2 + 2} \right] = 2 \left[ \frac{1}{4} - \frac{1}{8} \right]$ For $r=3$: $2 \left[ \frac{1}{3^2 - 3 + 2} - \frac{1}{3^2 + 3 + 2} \right] = 2 \left[ \frac{1}{8} - \frac{1}{14} \right]$ ... For $r=20$: $2 \left[ \frac{1}{20^2 - 20 + 2} - \frac{1}{20^2 + 20 + 2} \right] = 2 \left[ \frac{1}{382} - \frac{1}{422} \right]$

When we sum these terms, the intermediate terms cancel out: $S_{20} = 2 \left[ \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{8}\right) + \left(\frac{1}{8} - \frac{1}{14}\right) + \dots + \left(\frac{1}{382} - \frac{1}{422}\right) \right]$ The sum simplifies to: $S_{20} = 2 \left[ \frac{1}{2} - \frac{1}{422} \right]$

Step 5: Simplify the Sum to Find m/n $S_{20} = 2 \left[ \frac{211}{422} - \frac{1}{422} \right]$ $S_{20} = 2 \left[ \frac{211 - 1}{422} \right]$ $S_{20} = 2 \left[ \frac{210}{422} \right]$ $S_{20} = \frac{420}{422}$ Now, we need to simplify this fraction to its lowest terms, ensuring m and n are coprime. $S_{20} = \frac{420 \div 2}{422 \div 2} = \frac{210}{211}$ So, we have $\frac{m}{n} = \frac{210}{211}$. We need to check if $m=210$ and $n=211$ are coprime. 211 is a prime number. Since 210 is not a multiple of 211, they are coprime.

Step 6: Calculate m + n We are given that the sum is $\frac{m}{n}$, where m and n are coprime. We found $m=210$ and $n=211$. $m + n = 210 + 211 = 421$

Common Mistakes & Tips

• Incorrect Denominator Factoring: Ensure the difference of squares is correctly applied to the denominator. Mistakes in algebraic manipulation here will propagate.
• Identifying the Telescoping Pattern: Carefully match the second part of one term with the first part of the next term. Writing out the first 2-3 terms and the last term is very helpful.
• Coprime Check: Always simplify the final fraction $\frac{m}{n}$ to ensure $m$ and $n$ have no common factors before calculating $m+n$.

Summary The problem required summing a series where the general term could be decomposed into a difference of two functions. By manipulating the denominator of the general term to reveal a difference of squares and then using partial fraction decomposition, we expressed the general term as $2 \left[ \frac{1}{r^2 - r + 2} - \frac{1}{r^2 + r + 2} \right]$. This form allowed for the series to telescope. Summing the first 20 terms resulted in the fraction $\frac{210}{211}$. With $m=210$ and $n=211$ being coprime, their sum $m+n$ is $210+211=421$.

Final Answer The final answer is $\boxed{421}$ which corresponds to option (B).