# Finite Summation of Fractional Series

I have a finite summation series of identical fractions e.g $\frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{5}{3}$. Now lets say I add one to the denominator for the first two values of the series. I now have $\frac{1}{4} + \frac{1}{4} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{2}$. Now I do it again but this time to the first three values of the series. $\frac{1}{5} + \frac{1}{5} + \frac{1}{4} + \frac{1}{3} + \frac{1}{3} = \frac{79}{60}$. To complicate things again I add one to the denominator of the last three values of the series too. $\frac{1}{5} + \frac{1}{5} + \frac{1}{5} + \frac{1}{4} + \frac{1}{4} = \frac{11}{10}$

Is there an easy direct method to figure out the final answer $\frac{11}{10}$ given that we know that we did $[2,3]$ additions to the left of the series, and $[3]$ additions to the right of the series?

-

Well, lets say we are summing $n$ fractions $\frac{p}{q}$ and add $x$ to the denominators of $m$ terms.
The sum of the original series is $s = \frac{np}{q}$.
The sum of the new series is $s' = \frac{np}{q} - \frac{mp}{q} + \frac{mp}{q + x} = (m - n)\frac{p}{q} + \frac{mp}{q + x} = \frac{m - n}{n}s + \frac{mp}{q + x}$.
Of course I am making the assumption $m \leq n$. –  Alex Becker Mar 24 '11 at 4:52