The sum of the first 10 terms of the series $$\frac{1}{1 + 1^4 \cdot 4} + \frac{2}{1 + 2^4 \cdot 4} + \frac{3}{1 + 3^4 \cdot 4} + \cdots$$ is $$\frac{m}{n}$$, where $$\gcd(m, n) = 1$$. Then $$m + n$$ is equal to :

Let the $$k^{\text{th}}$$ term of the given series be

$$T_k=\frac{k}{1+4k^{4}},\qquad k=1,2,\dots,10.$$

1. Factor the denominator.
Observe that

$$4k^{4}+1=(2k^{2}-2k+1)\,(2k^{2}+2k+1).$$

2. Express $$T_k$$ as a difference of two fractions.
Using the identity above, write

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

Verification: $$\dfrac{1}{4}\left(\dfrac{1}{a}-\dfrac{1}{b}\right) =\dfrac{1}{4}\dfrac{b-a}{ab} =\dfrac{1}{4}\dfrac{(2k^{2}+2k+1)-(2k^{2}-2k+1)}{ab} =\dfrac{1}{4}\dfrac{4k}{ab} =\dfrac{k}{ab}.$$

3. Recognise the telescoping pattern.
Note that

$$2(k+1)^{2}-2(k+1)+1=2k^{2}+2k+1.$$

Thus if we define $$A_k=2k^{2}-2k+1,$$ then $$T_k=\dfrac14\left(\dfrac1{A_k}-\dfrac1{A_{k+1}}\right).$$

4. Sum the first 10 terms.

$$S_{10}=\sum_{k=1}^{10}T_k =\frac14\sum_{k=1}^{10}\left(\frac1{A_k}-\frac1{A_{k+1}}\right) =\frac14\left(\frac1{A_1}-\frac1{A_{11}}\right).$$

5. Evaluate the boundary terms.
$$A_1=2(1)^2-2(1)+1=1,$$
$$A_{11}=2(11)^2-2(11)+1=242-22+1=221.$$

Therefore

$$S_{10}=\frac14\left(1-\frac1{221}\right) =\frac14\cdot\frac{221-1}{221} =\frac14\cdot\frac{220}{221} =\frac{55}{221}.$$

6. Final step.
The reduced fraction is $$\dfrac{m}{n}=\dfrac{55}{221}$$ with $$\gcd(55,221)=1.$$
Hence $$m+n=55+221=276.$$

Option C which is: $$276$$