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 ___________

To solve the summation problem, we first need to simplify the general term of the series. Let the general term be $$T_r$$.

$$T_r = \frac{r}{r^4 + r^2 + 1}$$

Step 1: Factor the denominator

We can rewrite the denominator by adding and subtracting $$r^2$$:

$$r^4 + r^2 + 1 = (r^4 + 2r^2 + 1) - r^2$$ $$r^4 + r^2 + 1 = (r^2 + 1)^2 - r^2$$

Using the difference of squares formula, $$a^2 - b^2 = (a - b)(a + b)$$:

$$r^4 + r^2 + 1 = (r^2 - r + 1)(r^2 + r + 1)$$

So, the general term becomes:

$$T_r = \frac{r}{(r^2 - r + 1)(r^2 + r + 1)}$$

Step 2: Express as partial fractions

Notice that the difference between the two factors in the denominator is:

$$(r^2 + r + 1) - (r^2 - r + 1) = 2r$$

We can multiply the numerator and denominator of $$T_r$$ by $$2$$:

$$T_r = \frac{1}{2} \left[ \frac{2r}{(r^2 - r + 1)(r^2 + r + 1)} \right]$$

Now, substitute $$2r$$ with the difference of the factors:

$$T_r = \frac{1}{2} \left[ \frac{(r^2 + r + 1) - (r^2 - r + 1)}{(r^2 - r + 1)(r^2 + r + 1)} \right]$$

Splitting the fraction, we get a telescoping form:

$$T_r = \frac{1}{2} \left[ \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right]$$

Step 3: Evaluate the sum

Now we evaluate the sum from $$r = 1$$ to $$r = 25$$:

$$S = \sum_{r=1}^{25} T_r$$

Let's write out the first few terms:

For $$r = 1$$: $$T_1 = \frac{1}{2} \left[ \frac{1}{1} - \frac{1}{3} \right]$$

For $$r = 2$$: $$T_2 = \frac{1}{2} \left[ \frac{1}{3} - \frac{1}{7} \right]$$

For $$r = 3$$: $$T_3 = \frac{1}{2} \left[ \frac{1}{7} - \frac{1}{13} \right]$$

...

For $$r = 25$$: $$T_{25} = \frac{1}{2} \left[ \frac{1}{25^2 - 25 + 1} - \frac{1}{25^2 + 25 + 1} \right]$$

When we add all these terms together, all the intermediate fractions cancel out (telescoping series):

$$S = \frac{1}{2} \left[ 1 - \frac{1}{25^2 + 25 + 1} \right]$$

Calculate the value of the remaining denominator:

$$25^2 + 25 + 1 = 625 + 25 + 1 = 651$$

So, the sum is:

$$S = \frac{1}{2} \left[ 1 - \frac{1}{651} \right]$$ $$S = \frac{1}{2} \left[ \frac{650}{651} \right]$$ $$S = \frac{325}{651}$$

Step 4: Find $$p + q$$

We are given that the sum equals $$\frac{p}{q}$$, where $$\gcd(p, q) = 1$$. Let's check if the fraction $$\frac{325}{651}$$ is in its simplest form.

• Factors of $$325 = 25 \times 13 = 5^2 \times 13$$
• Factors of $$651 = 3 \times 217 = 3 \times 7 \times 31$$

Since there are no common factors, they are co-prime, meaning $$\gcd(325, 651) = 1$$.

Thus, $$p = 325$$ and $$q = 651$$.

Finally, calculate $$p + q$$:

$$p + q = 325 + 651 = 976$$

The final answer is 976.

$${p}/{q} = \frac{325}{651}$$, and $$p + q = 325 + 651 = 976$$.