Question

If the sum of the first 10 terms of the series \[ \frac{4 \cdot 1}{1 + 4 \cdot 1^4} + \frac{4 \cdot 2}{1 + 4 \cdot 2^4} + \frac{4 \cdot 3}{1 + 4 \cdot 3^4} + \ldots \] is \(\frac{m}{n}\), where \(\gcd(m, n) = 1\), then \(m + n\) is equal to _____.

Hint:

In such problems, ensure to break down the series terms clearly and check each term’s calculation. This helps in summing the terms and eventually determining \( m + n \).

Answer 1

Correct Answer: 441

Calculate the sum of the initial 10 terms of the series

\[ S_{10}=\sum_{n=1}^{10}\frac{4n}{1+4n^{4}}. \]

Concept Used:

Factor the denominator's quartic expression and employ partial fractions to establish a telescoping series.

\[ 1+4n^{4}=(2n^{2}-2n+1)(\,2n^{2}+2n+1\,). \]

Consequently,

\[ \frac{4n}{1+4n^{4}} =\frac{4n}{(2n^{2}-2n+1)(2n^{2}+2n+1)} =\frac{1}{2n^{2}-2n+1}-\frac{1}{2n^{2}+2n+1}. \]

Step-by-Step Solution:

Step 1: Restate each term using the aforementioned identity and aggregate from \(n=1\) to \(10\):

\[ S_{10}=\sum_{n=1}^{10}\left(\frac{1}{2n^{2}-2n+1}-\frac{1}{2n^{2}+2n+1}\right). \]

Step 2: Identify the telescoping property: the negative term corresponding to \(n\) cancels the positive term corresponding to \(n+1\), due to

\[ 2(n+1)^{2}-2(n+1)+1=2n^{2}+2n+1. \]

Step 3: Following the cancellation, only the initial positive term and the final negative term remain:

\[ S_{10}=\frac{1}{2\cdot1^{2}-2\cdot1+1}-\frac{1}{2\cdot10^{2}+2\cdot10+1} =1-\frac{1}{221} =\frac{220}{221}. \]

Final Computation & Result

Given \( \dfrac{m}{n}=\dfrac{220}{221} \) in its simplest form, the sum \( m+n \) is \( 220+221=\boxed{441} \).