Isn't is the sum of differences ?

Notice that, $r^{4}+r^{2}+1 = r^{4}+2r^{2}+1 - r^{2} = (r^{2}+1)^{2} - r^{2} \\ = (r^{2}+r+1)(r^{2}-r+1)$ $\mathrm{}$ Again, $r^{4}+r = r(r^{3}+1) = r(r+1)(r^{2}-r+1)$

Thus we get, $\sum_{r=1}^{n} \frac{r^{4}+r^{2}+1}{r^{4}+r} = \sum_{r=1}^{n} \frac{(r^{2}+r+1)(r^{2}-r+1)}{r(r+1)(r^{2}-r+1)}$ $= \sum_{r=1}^{n} \frac{(r^{2}+r+1)}{r(r+1)} = \sum_{r=1}^{n}\left ( 1+\frac{1}{r(r+1)} \right )$ $= \sum_{r=1}^{n} 1 + \sum_{r=1}^{n} \frac{1}{r(r+1)} = n + \sum_{r=1}^{n}\left ( \frac{1}{r} - \frac{1}{r+1} \right )$

Now, $\sum_{r=1}^{n}\left ( \frac{1}{r} - \frac{1}{r+1} \right ) = \left ( \frac{1}{1} - \frac{1}{2} \right )+\left ( \frac{1}{2} - \frac{1}{3} \right ) + ... + \left ( \frac{1}{n} - \frac{1}{n+1} \right )$ $=1-\frac{1}{n+1} \left [ \textrm{Telescoping} \right ]$

$\therefore n + \sum_{r=1}^{n}\left ( \frac{1}{r} - \frac{1}{r+1} \right ) = n + \left ( 1-\frac{1}{n+1} \right ) = \frac{675}{26} = 26 - \frac{1}{26}$ $\Rightarrow n-\frac{1}{n+1} = 25 - \frac{1}{26}$ $\therefore$ Its easy to see that $n = \boxed{25}$ satisfies the equation.

Python:

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from fractions import Fraction as frac
memo = {1:frac(3,2)}
def f(n):
    if n in memo:
        return memo[n]
    memo[n] = f(n-1)+frac(n**4+n**2+1,n**4+n)
    return memo[n]
n = 1
goal = frac(675,26)
while f(n) != goal:
    n += 1
print "Answer:", n