An algebra problem by Darel Gunawan

The problem can be written as $\displaystyle{\sum_{i=2}^n \frac{1}{x^3-x}}$ .

Observe, that $\displaystyle{\frac{1}{x^3-x} = \frac{1}{x(x+1)(x-1)}}$ . We can now do partial fraction decomposition and split this fraction into: $\displaystyle{ \frac{1}{x^3-x} = -\frac{1}{x} + \frac{1}{2} \left(\frac{1}{x+1} + \frac{1}{x-1}\right)}$

Now the problem is equivalent to $- \displaystyle{\sum_{i=2}^n \frac{1}{x}} + \frac{1}{2} \cdot \left(\displaystyle{\sum_{i=2}^n \frac{1}{x+1}} + \displaystyle{\sum_{i=2}^n \frac{1}{x-1}} \right)$ .

Consider $\displaystyle{\sum_{i=2}^n \frac{1}{x}}$ . Let this sum be equal to $z$ .

Observe that $\displaystyle{\sum_{i=2}^n \frac{1}{x+1}} = z - \frac{1}{2} + \frac{1}{n+1}$ and that $\displaystyle{\sum_{i=2}^n \frac{1}{x-1}} = z + 1 - \frac{1}{n}$ .

Now we can write: $\displaystyle{\sum_{i=2}^n \frac{1}{x^3-x}} = -z + \frac{1}{2}\left(2z + \frac{1}{m+1} - \frac{1}{m} + 1 - \frac{1}{2} \right) = \frac{1}{2} \left(\frac{1}{m+1} - \frac{1}{m} + \frac{1}{2}\right)$ .

Now we find the common denominator and do some factoring to get: $\boxed{\displaystyle{\sum_{i=2}^n \frac{1}{x^3-x}} = \frac{(m+2)(m-1)}{4m(m+1)}}$ . When $m = 100$ the fraction evaluates to $\frac{5049}{20200}$ . So our final solution is $5049 + 20200 = 25249$ .