The problem asks for the sum of the first 20 terms of a given series and then to calculate $m+n$ based on the sum represented as $\frac{m}{n}$ where $m$ and $n$ are coprime.
Series Term Analysis
The general term of the series is given by:
$ T_k = \frac{4k}{4+3k^2+k^4} $
Let's analyze the denominator $ D = k^4 + 3k^2 + 4 $. We can rewrite this using algebraic manipulation:
$ D = (k^4 + 4k^2 + 4) - k^2 $
Recognizing the perfect square $(k^2+2)^2$, we get:
$ D = (k^2+2)^2 - k^2 $
This is a difference of squares, $a^2 - b^2 = (a-b)(a+b)$, where $a = k^2+2$ and $b=k$. So:
$ D = ((k^2+2) - k)((k^2+2) + k) = (k^2 - k + 2)(k^2 + k + 2) $
Now, the general term $T_k$ becomes:
$ T_k = \frac{4k}{(k^2 - k + 2)(k^2 + k + 2)} $
Partial Fraction Decomposition for Telescoping Sum
We aim to express $T_k$ in the form $2 \left( \frac{1}{f(k-1)} - \frac{1}{f(k)} \right)$ for some function $f(k)$. Let $f(k) = k^2 + k + 2$. Then $f(k-1) = (k-1)^2 + (k-1) + 2 = k^2 - 2k + 1 + k - 1 + 2 = k^2 - k + 2$.
Consider the difference $f(k) - f(k-1)$:
$ f(k) - f(k-1) = (k^2 + k + 2) - (k^2 - k + 2) = 2k $
The numerator of $T_k$ is $4k$, which is $2 \times (2k)$. So we can write:
$ T_k = \frac{2(f(k) - f(k-1))}{f(k-1)f(k)} $
Separating the terms:
$ T_k = 2 \left( \frac{f(k)}{f(k-1)f(k)} - \frac{f(k-1)}{f(k-1)f(k)} \right) = 2 \left( \frac{1}{f(k-1)} - \frac{1}{f(k)} \right) $
Substituting back $f(k)$ and $f(k-1)$:
$ T_k = 2 \left( \frac{1}{k^2 - k + 2} - \frac{1}{k^2 + k + 2} \right) $
Calculating the Sum of the First 20 Terms
The sum of the first 20 terms, $S_{20}$, is given by $\sum_{k=1}^{20} T_k$. This is a telescoping series:
$ S_{20} = \sum_{k=1}^{20} 2 \left( \frac{1}{k^2 - k + 2} - \frac{1}{k^2 + k + 2} \right) $
Let's write out the terms:
- For $k=1$: $ 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 $k=2$: $ 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 $k=3$: $ 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 $k=20$: $ 2 \left( \frac{1}{20^2-20+2} - \frac{1}{20^2+20+2} \right) = 2 \left( \frac{1}{382} - \frac{1}{422} \right) $
Notice that the second part of each term cancels with the first part of the next term. The sum simplifies to:
$ S_{20} = 2 \left( \frac{1}{f(0)} - \frac{1}{f(20)} \right) $
Where $f(k) = k^2 + k + 2$. We need $f(0)$ and $f(20)$.
- $ f(0) = 0^2 + 0 + 2 = 2 $
- $ f(20) = 20^2 + 20 + 2 = 400 + 20 + 2 = 422 $
Substituting these values:
$ S_{20} = 2 \left( \frac{1}{2} - \frac{1}{422} \right) $
$ S_{20} = 2 \left( \frac{422 - 2}{2 \times 422} \right) = 2 \left( \frac{420}{844} \right) = \frac{420}{422} $
Simplifying the fraction by dividing the numerator and denominator by 2:
$ S_{20} = \frac{210}{211} $
Determining m, n, and m + n
The sum is given as $ \frac{m}{n} $, where $m$ and $n$ are coprime. We found the sum to be $ \frac{210}{211} $.
- $ m = 210 $
- $ n = 211 $
To check if they are coprime, we note that 211 is a prime number. Since 210 is not a multiple of 211, $m$ and $n$ are coprime.
The question asks for the value of $m+n$:
$ m + n = 210 + 211 = 421 $
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