If $$\sum_{r=1}^{25}\left( \frac{r}{r^{4}+r^{2}+1} \right)=\frac{p}{q},$$ where p and q are positive integers such that gcd(p,q)=1, then p+q is equal to ___________

The given sum is $$\sum_{r=1}^{25} \frac{r}{r^4 + r^2 + 1}$$.

First, factorize the denominator. Notice that $$r^4 + r^2 + 1 = (r^4 + 2r^2 + 1) - r^2 = (r^2 + 1)^2 - r^2 = (r^2 - r + 1)(r^2 + r + 1)$$.

Thus, the general term is $$T_r = \frac{r}{(r^2 - r + 1)(r^2 + r + 1)}$$.

Decompose $$T_r$$ using partial fractions. Assume:

$$\frac{r}{(r^2 - r + 1)(r^2 + r + 1)} = \frac{A r + B}{r^2 - r + 1} + \frac{C r + D}{r^2 + r + 1}$$.

Multiplying both sides by $$(r^2 - r + 1)(r^2 + r + 1)$$ gives:

$$r = (A r + B)(r^2 + r + 1) + (C r + D)(r^2 - r + 1)$$.

Expanding and equating coefficients:

For $$r^3$$: $$A + C = 0$$,
For $$r^2$$: $$A + B - C + D = 0$$,
For $$r$$: $$A + B + C - D = 1$$,
For constant term: $$B + D = 0$$.

Solving these equations:

From $$A + C = 0$$, $$C = -A$$.
From $$B + D = 0$$, $$D = -B$$.
Substituting into the $$r^2$$ equation: $$A + B - (-A) + (-B) = 0 \Rightarrow 2A = 0 \Rightarrow A = 0$$.
Then $$C = 0$$.
Substituting into the $$r$$ equation: $$0 + B + 0 - (-B) = 1 \Rightarrow 2B = 1 \Rightarrow B = \frac{1}{2}$$.
Then $$D = -\frac{1}{2}$$.

Thus, $$T_r = \frac{\frac{1}{2}}{r^2 - r + 1} + \frac{-\frac{1}{2}}{r^2 + r + 1} = \frac{1}{2} \left( \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right)$$.

Define $$f(r) = \frac{1}{r^2 - r + 1}$$. Then $$f(r+1) = \frac{1}{(r+1)^2 - (r+1) + 1} = \frac{1}{r^2 + r + 1}$$.

So, $$T_r = \frac{1}{2} \left( f(r) - f(r+1) \right)$$.

The sum becomes:

$$\sum_{r=1}^{25} T_r = \frac{1}{2} \sum_{r=1}^{25} \left( f(r) - f(r+1) \right)$$.

This is a telescoping series. Writing the terms:

$$\sum_{r=1}^{25} \left( f(r) - f(r+1) \right) = \left[ f(1) - f(2) \right] + \left[ f(2) - f(3) \right] + \cdots + \left[ f(25) - f(26) \right] = f(1) - f(26)$$.

Thus, the sum is $$\frac{1}{2} \left( f(1) - f(26) \right)$$.

Now compute:

$$f(1) = \frac{1}{1^2 - 1 + 1} = \frac{1}{1} = 1$$,
$$f(26) = \frac{1}{26^2 - 26 + 1} = \frac{1}{676 - 26 + 1} = \frac{1}{651}$$.

So, the sum is $$\frac{1}{2} \left( 1 - \frac{1}{651} \right) = \frac{1}{2} \left( \frac{650}{651} \right) = \frac{650}{1302}$$.

Simplify the fraction:

$$\frac{650}{1302} = \frac{650 \div 2}{1302 \div 2} = \frac{325}{651}$$.

Check if $$325$$ and $$651$$ are coprime. Factorize:

$$325 = 5^2 \times 13$$,
$$651 = 3 \times 7 \times 31$$.

No common prime factors, so gcd is 1. Thus, $$\frac{p}{q} = \frac{325}{651}$$, and $$p + q = 325 + 651 = 976$$.