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

We need the sum of the first 10 terms of the series

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

Concept Used:

Factor the quartic in the denominator and use partial fractions to obtain a telescoping series.

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

Hence

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

Step-by-Step Solution:

Step 1: Rewrite each term using the identity above and sum 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: Observe telescoping: the negative term for \(n\) cancels the positive term for \(n+1\), since

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

Step 3: After cancellation only the first positive and the last negative terms 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

With \( \dfrac{m}{n}=\dfrac{220}{221} \) in lowest terms, \( m+n=220+221=\boxed{441} \).

Answer 2

The general term \( T_r \) of the series is given by: \[ T_r = \frac{4r}{1 + 4r^4} \] Hence, for the first few terms: \[ T_1 = \frac{4}{1 + 4 \times 1^4} = \frac{4}{5} \] \[ T_2 = \frac{4}{(2^2 + 2 \times 2 + 1)} = \frac{4}{13} \] \[ T_3 = \frac{4}{(3^2 + 3 \times 2 + 1)} = \frac{4}{21} \] \[ \vdots \] Now, the sum of the first 10 terms is given by: \[ S_{10} = T_1 + T_2 + \cdots + T_{10} \] Thus, for \( m \) and \( n \), we find: \[ S_{10} = \frac{1}{181} + \frac{220}{221} = \frac{220}{221} \] Hence, \( m + n = 220 + 221 = 441 \).