General Term Analysis

The series is given by \( \frac{1}{1+1^{4} \times 4} + \frac{2}{1+2^{4} \times 4} + \frac{3}{1+3^{4} \times 4} + \dots \). The general term, \( T_k \), is:

\( T_k = \frac{k}{1+4k^4} \)

Factor the denominator using the identity \( 1+4x^4 = (1+2x^2+2x)(1+2x^2-2x) \). Substituting \( x=k \):

\( 1+4k^4 = (2k^2 + 2k + 1)(2k^2 - 2k + 1) \)

We express \( T_k \) using the difference of two fractions. Note that \( (2k^2 + 2k + 1) - (2k^2 - 2k + 1) = 4k \).

\( T_k = \frac{1}{4} \times \frac{4k}{(2k^2 - 2k + 1)(2k^2 + 2k + 1)} \) \( T_k = \frac{1}{4} \left( \frac{1}{2k^2 - 2k + 1} - \frac{1}{2k^2 + 2k + 1} \right) \)

Telescoping Series Summation

Define \( f(k) = 2k^2 - 2k + 1 \). Then \( f(k+1) = 2(k+1)^2 - 2(k+1) + 1 = 2(k^2+2k+1) - 2k - 2 + 1 = 2k^2 + 2k + 1 \).

The general term simplifies to a telescoping form: \( T_k = \frac{1}{4} \left( \frac{1}{f(k)} - \frac{1}{f(k+1)} \right) \)

The sum of the first 10 terms, \( S_{10} \), is: \( S_{10} = \sum_{k=1}^{10} T_k = \frac{1}{4} \sum_{k=1}^{10} \left( \frac{1}{f(k)} - \frac{1}{f(k+1)} \right) \) This sum telescopes to: \( S_{10} = \frac{1}{4} \left( \frac{1}{f(1)} - \frac{1}{f(11)} \right) \)

Calculating the Sum \( \frac{m}{n} \)

First, compute \( f(1) \) and \( f(11) \): \( f(1) = 2(1)^2 - 2(1) + 1 = 1 \) \( f(11) = 2(11)^2 - 2(11) + 1 = 2(121) - 22 + 1 = 242 - 21 = 221 \)

Now, calculate \( S_{10} \): \( S_{10} = \frac{1}{4} \left( \frac{1}{1} - \frac{1}{221} \right) = \frac{1}{4} \left( \frac{221 - 1}{221} \right) = \frac{1}{4} \times \frac{220}{221} = \frac{55}{221} \)

So, \( \frac{m}{n} = \frac{55}{221} \).

Finding \( m+n \)

We have \( m = 55 \) and \( n = 221 \). Check for the greatest common divisor (GCD): \( m = 55 = 5 \times 11 \) \( n = 221 = 13 \times 17 \)

The prime factors of \( m \) and \( n \) are distinct, so \( \operatorname{gcd}(55, 221) = 1 \).

The value required is \( m+n \): \( m+n = 55 + 221 = 276 \)